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NEW PLANE AND SPHERICAL 
TRIGONOMETRY, SURVEY- 
ING, AND NAVIGATION 



BY 

G. A. WENTWORTH 

Author of a Series of Text-Books in Mathematics 



TEACHERS' EDITION 



BOSTON, U.S.A. 
GINN & COMPANY, PUBLISHERS 

. . 1903 



THE LIBRARY OF 
CONGRESS, J 


Two Copies Receive** 1 


SEP 22 1903 


Copyright tntry J 
CLASK Co XXc No 


COPY B. 



4? i 



ro 



y 



Copyright, 1903 
By GEORGE WENTWORTH 



ALL RIGHTS RESERVED 



< • 



x> 



PREFACE 

This edition is intended for, teachers, and for them only. 
The publishers will under no circumstances sell the book 
except to teachers of Went worth's Trigonometry ; and every 
teacher must consider himself in honor bound not to leave his 
copy where pupils can have access to it, and not to sell his 
copy except to the publishers, Messrs. Ginn & Company. 

It is hoped that young teachers will derive great advantage 
from studying the systematic arrangement of the work, and 
that all teachers who are pressed for time will find great 
relief by not being obliged to work out every problem. 

G. A. WENTWORTH. 



PLANE TRIGONOMETRY 



TEACHERS' EDITION. 



Exercise I. Page 2. 



1. Reduce the following angles to 
circular measure, expressing the 
results as fractions of it : 60°, 45°. 
160°, 195°, 11° 15', 123° 45'. 37° 30 7 . 

00° — - G -°- it — ] -it 
45° — - 4 5 - it — l tt 

150° = HJ 2T = g It. 

195° — 19a it — 13 7T 

Hi 1 

11° 15' = ill* =— tt. 

180 16 

1loor/ 123J 11 

1*23° 4o = — — ?r = — it. 
180 16 



37 



■30-^* =^n. 



180 



24 



2. How many degrees are there 
in | n radians ? } it radians ? j it 
radians? \\ 7t radians? T t- tc radians? 

§ n radians = f of 180° = 120°. 

\ n radians = | of 180° = 135°. 

\ n radians = | of 180° = 112° 30'. 
if it radians = \\ of 180°= 168° 45'. 
T V it radians = ^ of 180°= 84°. 

3. What decimal part of a radian 
is 1°? r? 



V = 



n 3.1416 

: radian = 

180 180 

: 0.017453 radian. 

0.017453 



radian 



60 



radian 



= 0.0002909 radian. 

4. How many seconds in a radian ? 
1 radian = 57° 17' 45" 

= 206. 265". 

5. Express in radians one of the 
interior angles of a regular octagon ; 
of a regular dodecagon. 

The sum of all the interior angles 
of a regular octagon is 8 x 180° — 
360° = 8* -2?t = e>n. Hence, each 
interior angle 

3 



— -it radians : 



- it radians. 



The sum of all the interior angles 
of a regular dodecagon is 12 x 180° 
- 360° = 12 it - 2 it = 10 7t. Hence, 
each interior angle 

= — tc radians = - it radians. 

12 6 



PLANE TRIGONOMETRY. 



6. On the circumference of a cir- 
cle of 50 feet radius an arc of 10 feet 
is laid off. How many degrees in 
the angle at the centre by this arc ? 

It subtends if radian = i radian 



57° 17' 45" 



11° 27' 33" 



7. The earth's equatorial radius 
is approximately 3963 miles. If two 
points on the equator are 1000 miles 
apart, what- is their difference in 
longitude ? 

Their difference in longitude is 
Hii radian, or 

D 17' 45 



2 00 0. 

3 9 6 3 



= 14° 27' 28" 



8. If the difference in longitude 
of two points on the equator is 1°, 
what is the distance between them 
in miles ? 

By Example 3, 1° = 0.017453 
radian. Hence, 1 Q on the equator 
is equal to 0.017453 x 3963 miles = 
69.166 miles. 

9. What is the radius of a circle 
if an arc of 1 foot subtends an angle 
of 1° at the centre ? 

Since 1° of arc = 1 foot, 1 radian 



3.55 inches = the radius. 



or 57° 17' 45" = 57^ feet = 57 feet 



10. In how many hours is a point 
on the equator carried by the rota- 
tion of the earth on its axis through a 
distance equal to the earth's radius ? 

The earth turns through 360° 
= 2 Tt radians in 24 hours. Hence, 

24 

it turns through 1 radian in — 

12 2n 

hours = hours = 3 hours 49 

3.1416 
minutes 11 seconds. 

11. The minute hand of a clock 
is 3J feet long. How far does its 
extremity move in 25 minutes ? 
[Take n = - 2 7 2 -.] 

The circumference which is passed 
over in 60 minutes is 2 it x 8f feet. 
Hence, the arc passed over in 25 
minutes is 



\% of 2 n x 3i ft. 



= Vh x 2 x ¥ x 
= 91 ft. 



ft. 
9 ft. 2 in. 



12. A wheel makes 15 revolutions 
a second. How long does it take to 
turn through 4 radians ? [Take 

Tt = -V-.] 

The wheel turns through 2 Tt 
radians in T ^ of a second. Hence, 
it turns through 4 radians in 
4 x ( T *j second ~ 2rt)= T | 3 second. 



Exercise II. Page 5. 



1. What are the functions of the 
other acute angle B of the triangle 
AC B (Fig. 2)? 



sin B — 



cos B — 



tan B — 



cot 2? = ?. 



sec B — - • 
a 



esc B ■ 



T E A C H E E S ' E J> T T J O N . 



2. Compare the functions of A 
and B, and show that 

sin A = cos B. 
cos A — sin B, 
tan A — cot B, 
cot ^4 = tan J5, 
sec v4 = esc 2?, 
esc ^4 = sec B< 
vers ^4 = covers B. 
covers A = vers B. 



(H) 



cos J? = 
sin 2? = 
cot 5 = 
tan B = 
esc 7/ — 
sec B — 



vers ^4 = — - » covers 7> = — - 
c c 

. c-a c-( 

covers A = , vers B = — 

c c 



3. Find the values of the func- 
tions of A< if a, fr, c, respectively, 
have the following values : 
(i) 3, 4, 5. (iv) 9, 40, 41. 
(ii) 5, 12. 13. (v) 3.9, 8, 8.9. 
(Hi) 8, 15, 17. (vi) 1.19, 1.20, 1.09. 



sin .4 


c 


cos A 


_b 

~ c 


tan A 


a 

~b 


cot A 


b 


sec A 


t 
= b 


esc A 


c 




a 



(i) 



sin A = 
cos A — 

tan ^4 = 



eotyl = 
sec A = 
esc A — 



• A ° 

sin A c= — . 

13 


cot A ~ — 
5 


A 12 

cos A — — 
13 


13 

sec* ^4 = — 

12 


tan A — — . 
12 


A 13 

csc A = — 
5 



(iii) 



• A 8 

sin A — 

17 




* A W 

COt ^4 = — 

8 


A l5 

cos A — — 
17 




^ 17 

sec -d. e= — 

15 


Q 

tan A = — 
15 


(iv) 


A 17 

csc A — — 

8 


• a ° 
sin A — — • 

41 




I A & 

cot A = — 
9 


A 40 

cos A = — 
41 




^ 41 
sec 4= — 

40 


9 
tan A — — 
40 


(v) 


csc A : = — 
9 


• a » 

sin A = — 

89 




cot ^4 = — 
39 


A 80 

cos A = — 
89 




^ 89 
sec A = — 

80 


tan A — — 
80 




csc ^4 = — 
39 



(vi) 



sin ^4 


119 

~ 169* 


cot A 


_ 120 
" 119 


COS J- 


120 
~169' 


sec A 


_169 
~ 120 


tanJ. 


_119 
~120* 


csc A 


_169 
~ 119 



4 



PLANE TRIGONOMETRY. 



4. What condition must be ful- 
filled by the lengths of the three 
lines a, b, c (Fig. 2) in order to 
make them the sides of a right tri- 
angle ? Is this condition fulfilled 
in Example 3 ? 

The condition is a 2 + b 2 = c 2 . 

It is ; for 
32 + 42 = 52. 92 + 40 2 = 41 2 . 



5 2 -f 12 2 = 13 2 . 



3.9 2 + 8 2 = 8.9 2 . 



8 2 + 15 2 =17 2 . 1.19 2 -fl.20 2 =1.69 2 . 

5. Find the values of the func- 
tions of A, if a, 5, c, respectively, 
have the following values : 
(i) 2 mn, m 2 — n 2 , m 2 + n 2 . 

.... 2x2/ , x 2 + y 2 

(11) — , x + y, - - 

' x- 2/ x-2/ 

(iii) pgr, grs, rsp. 

(iv) — , — , — • 
pq sq ps 

(i) 





c 


m 2 + n 2 


COS ^i 


_ 6 _ 


m 2 — n 2 


m 2 + n 2 


tan J. 




2 wn- 


m 2 — n 2 


cot J. 


_ 6_ 
a 


m 2 — n 2 


2 mn 


sec A 


c 
~6 ~ 


?n 2 -f n 2 


m 2 — n 2 




c 


?n 2 -f n 2 



sin ^1 



2x?/ 



a 

(ii) 
x 



2 mn 



x 



x - y 
cos^L = (x+2/) x 



x 2 - 

X - 



2x2/ 
x 2 + 2/ 2 ' 



2/ 



E 2 - 



r 



.x 2 -f 2/ 2 x 2 + 2/ 2 



tan .A = x 



x - y x + 2/ 
x - 2/ 
2x2/ 
1 



cot ^1 = (x + y) x 

A x 2 + y 2 

sec A = — x 

x — 2/ x + y 

. x 2 -f y 2 x - y 

esc .4 = — x — — - 

x - y 2xy 

(iii) 

a PQ r Q 4 

,sm A = =-^- = - • cos^l = 
rsp s 

tan J. = — = -. cot^L = 
qrs s 

A TS P P A 

sec A == — - = — • esc A = 
qrs q 

(iv) 



2x?/ 


x 2 - 2/- 
x 2 - 2/ 2 


2x2/ 
x 2 + 2/ 2 


x 2 - 2/ 2 
* 2 + 2/ 2 



sm A = — x — = 
pq nr 

mv ps 

cos A = ■ — x — = 
sg nr 

mn set 
tan ^ = — x — - = 
pq mv 

CQtA = ™ x P! = 

sq mn 

- . nr sq 
sec A = — x — — = 
ps mv 

nr pq 
esc A = — x £1 ^- = 
ps mn 



2X2/ 

</rs _ </ 
rsp ~ p 

qrs _ s 

"pqr~ p 

rsp _ s 

pqr " g ' 

ms 
gr' 

mjpu 
nqr 
ns 
pv 

pv 

ns 

nqr 

mpv 

gr 

ms 



6. Prove that the values of a, 6, 
c in (i) and (ii), Example 5, satisfy 
the condition necessary to make 
them the sides of a right triangle. 

W 

a 2 + b 2 = c 2 . 
(2 mn) 2 + (?n 2 - n 2 ) 2 

= (?n 2 -f n 2 ) 2 . 
4 m 2 n 2 -f- m 4 - 2 m 2 n 2 + n 4 

= m 4 -}-2m 2 n 2 4-?i 4 . 
m 4 +2 ?n 2 n 2 + ?i 4 = ?n 4 + 2 m 2 n 2 -f n 4 . 



•TEACHEKS EDITION. 



(") 



\ x - y ) ' 



4iX 2y2 



x 2 - 2xy + y 2 



4- x 2 4- 2 xy + y- 
x 4 + 2x 2 y 2 + ?/ 1 



x 2 - 2 x?/ 4- */ 2 
4 x 2 ?/ 2 4- x 4 - 2 x 2 y 2 -f 2/ 4 

= x 4 4- 2 xV 2 4- yK 
x 4 4- 2 x 2 y 2 4- 2/ 4 = x 4 + 2 x 2 ?/ 2 4- 2/ 4 - 

7. What equations of condition 
must be satisfied by the values of 
a, 6, c in (iii) and (iv). Example 5. 
in order that the values may repre- 
sent the sides of a right triangle ? 

(iii) 

p'2g2 r 2 + g2 r 2 s 2 _ r 2 s 2p2 ? 

or p 2 q 2 4- g 2 s 2 = p 2 s 2 . 

(iv) 

??i 2 ri 2 ra 2 v 2 ?i 2 r 2 



p 2 ? 2 



+ ■ 



p^ 





c 


= V24 2 4- 143 2 
= V21025 






= 145. 




sin 


A 


_ 24 

7ui> = 


= cos B. 


cos 


A 


_143_ 
~ 145" 


- sin B. 


tan^l 


24 
~143 = 


- cot B. 


cot 


A 


143 

~ 24 " 


- tan B. 



or m 2 ri 2 s 2 4- m 2 p 2 v 2 — n 2 q 2 r 2 . 

8. Given a 2 + b 2 = c 2 ; find the | 
functions of >1 and B when a = 24, 
b = 143. 



sec A = = esc B. 

143 

A 145 I? 

csc vi = = sec B. 

24 



9. Given a 2 4- 6 2 = c 2 ; find the 
functions of A and 5 when 
a = 0.264, c = 0.265. 



b = V(c 4- a) (c - a ) 
- Vp.529 x .001 
= Vo. 000529 
= 0.023. 

. " a 264 
sm ^4 = - = — = cos B. 
265 



cos A 



tan A 



cot ^4 



c 

6 _ 23 

c ~ 265 

a _264 

^~^3" 

ft _ 23 

a~264 



: sin B. 



= cot B. 



tan 2*. 



. c 265 ■ 

sec -4. = - = — = csc B. 
b 23 



csc A 



265 
264 



sec B. 



10. Given a 2 4- b 2 = c 2 ; find the 
functions of A and B when 
6 =9.5, c = 19.3. 
a = V (c-4- 6) (c - a) 
= V28.8 x 9.8 
= V282.24 
= 16.8. 

• , a 168 

sm a = — = = cos 71 " 

c 193 

j 6 95 ' • ' » 
cos A = - = — - = sm is. 

c 193 

. a 1^8 

tan A — - = ^-=- = cot i>. 
6 95 

b 95 
cot J. = - = — r = tan B. 

a 168 



I'LANE TBIGOXOMETUY. 



e 193 

see A = - = -zzr = esc i>. 



cse J. 



95 

193 
168 



: sec B. 



11. Given a 2 4- b 2 =- c 2 ; find the 
functions of A and B when 

a = Vp 2 + g 2 ? 5 = V2 pq. 
c = Vj^-f p -f 2pg 



sin ^1 



a _ Vj> 2 -f ff 2 

" e ~ p 4- g 



6 V2pg 

COS A = - = ; 

c p 4-g 



: cos B. 



: sin J?. 



a vS 4- g 2 
tan A = - = ^_* = cot J5. 




cot ^4 = 



sec -4 = 



. c P + g 

csc *4 = - == 



a Vp 2 4- g 2 



= tan 2*. 



csc B. 



== sec J5. 



12. Given a 2 4- &* = c 2 ; And ilit 
functions of JL and i? when 

a = Vp 2 4- pg, c = p 4- g. 

52 sb C 2 _ a 2 

= g 2 + P9- 

.-. 5 = Vg 2 4. pg. 

a Vp2 4-pg 
sin -A = - = ■ - f = cos 7;. 



cos J. 



Vg 2 4- PQ 
P.+ g 

* . a ^P 2 ± PQ 

tan A = - = . .■..-: 

b Vg2 4- pq 



== sin 7>\ 



^v 



= * £ = cotB. 



cot A 



b _ Vg'2 _j_ p^ 

a V»2 



p J 4- pq 



4 



=z \± = tan B. 



sec ^4 == : 



p + q 



pq 



csc A = 



Vg2 + 
ft Vp2 _|_ pg 



-=8 csc B. 



c? sec J5. 



13. Given a 2 4- & 2 5= e 2 ; find the 
functions of A and J5 when 
b = 2 Vpg, c — p 4- g- 

a 2 = p2 4. 2pg 4- g 2 - 4pg 
-p 2 -2pg + q 2 . 
.-. a=p -q. 
a p -q 



sin Jl 



cos A 



i> ■ + 2 

2 Vp^ 



— cos J5. 



sin JB. 



tan B. 



b 2 y/pq 

b 2 V^ 

cot .A = - — 

a p-g 

2 Vpg 

C P 4" g r> 

csc A = - = " - * = sec J5. 
a p - q 



14. Given a 2 + 52 -- c 2 . fi n d the 
functions of A when a = 2 6. 

a = 26. 

a 2 4- & 3 = c* 

4 b 2 4- & 2 = c 2 . 

56 2 = c 2 . 

.-. c.= 6 V5. 



TEACHERS EDITIOK. 



sin A 



c 6 V5 
c 6 Vo 



tan A = - = — = 2. 
6 o 



cot A — 



1 

&V6 



vd /r 

sec ^4 = - = — -— = V 5. 
o 

esc A — - — — — - = J v 5. 
a 26 



15. Given a 2 4- W = c 2 ; find the 
functions of A when a = f c. 



6 = Vc 2 - a 2 



= Vc 2 - 
3 



4 r 2 
9 C 



Wi 






. • a | c 2 

sin ^4 = - = ^— = - • 
c c 3 

j b 3 
cos ^4 = - = 

c c 

tan^ = ? = -lL- = § V5. 
o c /r 

- Vo 

3 

cot J. = - = = I V 5. 

a §c 

sec .4 = - = = 4 Vs.- 

6 c /7 5 

3 V5 

- ■ . c c 3 
esc A = - = — = - • 
a § c 2 



16. Given a 2 + & 3 =-e* ; find the 
functions of A when a 4- 6 = f c. 

a -\-b-\c. 

a' 2 -\-b? = c 2 . 

a 2 + b' 2 + 2 a6 = H c* 

a 2 - 2 «6 -f ft 2 = t 7 ? A 

a - b = - v 7 7. 
4 



a -f 6 : 



§ 



c. 



sin JL 



2& = 5 C _<V7. 
4 4 



& = -(5-^ 7). 

8 V 

4 4 

a = -(5 4- V7). 



-(5 4- ^") /- 

a S K ' 5 + V7 



cos ^4 = 



6 8 



3 (5 - V7) 



5-V7 



a 5 4- ^7 16 4- 5 V7 

tan ^4 = - = — — — ; = 

h 5-V7 9 

' 6 5-V? 16-5V7 

cot A = - = ■ 



5 4- V? 



c 



sec ^4 = - 



6 c 



(o-Vl) 



.= f<6 + VT). 



esc ^ = - = — =|(ft- V7). 



PLANE TRIGONOMETRY. 



•'■ 17. Given a 2 .-f 6 2 = c 2 ; find the 
functions of A when a — b = Jc. 



a 2 - 2 ab 4- & 2 = 
a 2 



16 

+ 6 2 = c 2 . 



2ab 



_15c 2 
16 ' 

4- & 2 = c 2 . 



31 r 2 

a 2 + 2a6 + 6 2 = — • 
16 

a 4- & = - V31. 
4 

a — 6 = -• 
4 



2a 



: C -V3l + 1 

4 4 

c 



,..a = ^(V31 + l). 
4 4 



.-. & = -(V31 -1). 
a 8 V ' V31 + 1 



sin A = - = 

c c 8 

-(V31-1) ._ 

cos A = - = = — 

c c 8 



tan J. = 
cot A 



a _ V.31 4-1 _ 16+V31. 
& Vm. - 1 15 

b V31-1 lfr-V5T 



a V31 4-1 

A C 8 

sec A =. - = — — 

b V31 - 1 

= A(V31 + 1). 

A C 8 

csc A — - = — == 



15 



a V314-.1 
= T \(V31-1). 



18. Find a if sin A — f , and 
c = 20.5. 







sin ^1 = 


_a_3 
~c ~ 5 










a 
20J3~ 


3 
"5* 










.-. a - 


= 12.3. 








19. 


Find b if cos ^4 = 


0.44, 


and 


c 


= 3.5. 












A b 

cos JL = - = 
c 


= 0.44. 










6 

3^5 = 


^0.44. 










.«. 6: 


= 1.54. 








20. 


Find a if 


tan^l 


_ 11 

— ."3 ' 


and 


h 


= 2 


:V 









tan A = 

b 2 T \ 3 

11a 11 

27 ~~S' 

.-. a = 9. 

21. Find b if cot A = 4, and 
a = 17. 

b _b^ 
a ~~ 17 ~ 

.-. & = 68. 



cot A 



4. 



22. Find c if sec A = 2, and 

6 = 20. 

sec ^1 = - = — = 2. 

6 20 

,\ c = 40. 

23. Find c if csc A = 6.45, and 

a = 35.6. 

c c 

csc ^L = - = = 6.45. 

a 35.6 



, c = 229.62. 






teachers' edition. 



9 



24. Construct a right triangle, 
given c = 6, .tan A = f. 



tan ^L 



.-. tt:6 = 3:2. 

Draw .42* = 2. 
Draw BC ± to AB = 3. 
Join C and ^4. 

Prolong ,4 C to D, making ^4D=6. 
Draw BE 1. to AB produced. 
Rt. A AED is similar to rt. A 
ABC. 

.-. ABE is the rt. A required. 

25. Construct a right triangle, 
given a = 3.5, cos A = %. 

Construct rt. A A'B'C so that 
b' = 1, c' = 2. 

Then cos A = i. 

Construct A ABC similar to 
A'B'C, and having a = 3.5. 

Then ABC is rt. A required. 

26. Construct a right triangle. 

given 6 = 2, sin ^4 = 0.6. 

Construct rt. A A'B'C, making 
a' = 6, c' = 10. 

Then sin A' = T \. 

Construct A ABC similar to 
A'B'C, and having 6 = 2. 

Then ABC is rt. A required. 

27. Construct a right triangle, 
given 6 = 4, esc A = 4. 

Construct rt. A A'B'C, having 
<:' = 4, a' = 1. 

Then construct A ^4I?C similar 
to ^/.B'C", and having 6 = 4. 

Then ABC is rt. A required. 



28. In a right triangle, c = 2.5 
miles, sin A = 0.6, cos ^4 = 0.8 ; 
compute the legs. 

A b 

cos A = — 

c 

.-. 6 = c cos .1 
.-. 6 = 2. 



sin J. = - • 
c 

.-. a = c sin ^1. 
.-. a — 1.5. 



30. Find, by means of the table, 
the legs of a right triangle if A =20°, 
c = 1 ; also, if ^1 = 20°, c = 4. 



4 = 20°, 



c = l. 



sin J. = 



a 



.♦. a = csin^.. 
.-. a = 0.342. 

cos A = - • 
c 

.-. 6 = ccos^4. 

.-. b = 0.940. 

^ = 20°, c = 4. 

.-. a = 4 x 0.342 

= 1.368. 
.-. 6 = 4 x 0.940 

= 3.760. 

31. By dividing the length of a 
vertical rod by the length of its 
horizontal shadow, the tangent of 
the angle of elevation of the sun at 
the time of observation was found 
to be 0.82. How high is a tower, if 
the length of its horizontal shadow 
at the same time is 174.3 yards ? 

tan .4 =? = 0.82. 

b 

.-. a = 0.82 6. 

6 = 174.3 yards. 
.-. a = 0.82 of 174.3 yards 
= 142.926 yards. 



10 



PLANE TRIGONOMETRY. 



Exercise III. Page 9. * 



1. Represent by lines the func- 
tions of an acute angle larger than 
that shown in Fig. 3. 

n S 



/ N 


-\ 


V 


T 


/ 


\ 


A 


1 


i 


1 



Fig. 3. 

2. If x is an acute angle, show 
that sin x is less than tan x. 

In Fig. 3, OM.MP = OA:AT 
But OM < OA. 

r.MP<AT. 
.-. sinx<tanx. 

3. If x is an acute angle, show 
that sec x is greater than tan x. 

OT — sec x, AT — tan x. 
In it. A OA T, hyp. Or>side A T. 
.-. sec x > tan x. 

4. If x is an acute angle, show 
that esc x is greater than cot x. 

OS = esc x, BS = cot x. 
In rt. A BOS, hyp. OS > side BS. 
.-. CSCX>COtX. 

5. Construct the augle x if tan x 
= 3. 

Let (D BAB' be a unit circle, with 
centre O ; then construct A T tan- 
gent to the circle at A = 3 OA ; 
then A OT is required angle. 



6. Construct the angle x if esc x 
= 2. 

Let O ABA' be a unit circle, with 
centre O ; construct BS tangent to 
the circle at B = 2 OJL ; draw 06' ; 
then A OS is required angle. 

7. Construct the angle x if cos x 

= f 

Take OM = i radius OA . At J/ 
erect a _L to meet the circumference 
at P. Draw OP. 

Then POM is the angle required. 

8. Construct the angle x if sin x 
= cos X. 

Let MP — sin x and OM — cos x. 
But, by hypothesis, MP = Oif. 
.-. by Geometry, x == 45°. 
Hence, construct an Z. 45°. 

9. Construct the angle x if sin x 
= 2 cos x. 

Construct rt. Z PMO, making 
Pif = 2 OJf . Draw OP. 
Then POM" is the angle required. 

10. Construct the angle x if 
4 sin x = tan x. 

Take ± of radius OA to Jf. At 
3f erect a ± to meet the circum- 
ference at P. Draw OP. 

Then POM is the required angle. 

11. Show that the sine of an 
angle is equal to one-half the chord 
of twice the angle. 

Have given Z. POA. 

Construct POB = 2 POA. Draw 
chord PB. Then it is ± to OA ; 
and PJf, its half, is the sine of POA. 

.-. sinx = ^ chord 2x. 



TEACHERS EDITION. 



11 



12. Find x if sin x is equal to 
one-half the side of a regular in- 
scribed decagon. 

Let A C be a side of a decagon. 

Then — = 36° or ^.OC. 
10 

Draw OB bisecting AC. Then 
Z AOC is bisected, and Z AOB 
= 18°. 

But the sine of A OB = $ A C. 

.\xotAOB= 18°. 

13. Given x and y, x + y being 
less than 90° ; construct the value 
of sin (x + y) — sin x. 

Let AB = sin (x -h y) in a circle 
whose centre is O, and CD = sin x. 

Then, with a radius equal to CD. 
describe an arc from B, as centre, 
cutting AB at E. 

Then EA is the constructed value 
of sin (x -f y) — sin x. 

14. Given x and ?/, x + y being 
less than 90° ; construct the value 
of tan (x -f y) — sin (x + y) + tan x 

— sin x. 

Let ^4J3 = sin (x -f ?/), 

and CD = sin x ; 

also EF = tan (x -f ?/), 

and 6rF = tan x. 

From F with a radius = ^4i? take 
FH. 

From £T with a radius = 6rF take 
HI. 

From J in the opposite direction 
with a radius = CD take JJT. 

Then EK is the constructed value 
of tan (x -f y) — sin (x + y) + tan x 

— sin x. 

15. Given an angle x ; construct 
an angle y such that sin y = 2 sin x. 



Let ^4# be the sine of the Z x in 
a circle whose centre is 0. 

Draw AC perpendicular to the 
vertical diameter. 

Then CO = AB. 

Take CF on vertical diameter 
= CO. Draw FD perpendicular to 
vertical diameter, and meeting cir- 
cumference at D. 

Draw DE perpendicular to OB 
and draw OD. 

OF = 2 CO by construction. 

ED = FO; FO being the projec- 
tion of the radius OD. 

.-. DE = 2 AB, and DOB = angle 
required. 

16. Given an angle x ; construct 
an angle y such that cos y — \ cos x. 

Let 0B = cos AOB. 

Erect a _L CD at C, the middle 
point of OB, and meeting the cir- 
cumference at D. Draw DO. 

Then DOB is the angle required. 

17. Given an angle x ; construct 
an angle y such that tan y — 3 tan x. 

Let ^L B be the tangent of x. 

Prolong ^45 to C, making AC 
= 3 ^1J3, and draw OC from 0, the 
centre of the circle. 

CO A is the required angle. 

18. Given an angle x ; construct 
an angle y such that sec y = esc x. 

Since sec y = esc x, 

c _ c 
b ~ a' 

.-. a = b. 
Hence, construct an isosceles right 
triangle. 
The required angle will be 45°. 



12 



PLANE TRIGONOMETRY. 



19. Show by construction that 
2 sin A > sin 2 A. 

Construct Z BOC and Z COA 
each equal to the given Z A. 

Then J.I? = 2 sin ^, and AD, the 
± let fall from A to OB, = sin 2 ^4. 
But^LB>AZ). 

Hence, 2 sin .A > sin 2 A. 

20. Given two angles^, and jB, 
A + B being less than 90° ; show 
that sin (A + B) < sin ^4 -f sin B. 

Construct HOK = ZA, and COR 
= ZB. 

Then sin (4 + B) = CP, sin A 
= jOT^T, sin J5 = CD. 

Now CP<CD + DE, 

and UK>DE. 



.. CP<CD + HK. 

.-. sin (.A + £) < sin A. + sin JB. 

21. Given sinx in a unit circle; 
find the length of a line correspond- 
ing in position to sin x in a circle 
whose radius is r. 

1 : r = sin x : the required line. 
.-. length of line required = rsin x. 

22. In a right triangle, given the 
hypotenuse c, and also sin A = m, 
cos A = n ; find the legs. 

a 

c 

= cm. 
_6 
~ c 
.-. 6 = en. 



sin A = 



,\ a : 



cos A 



■■ m. 



: n. 



Exercise IV. Page 12. 



1. Express as functions of the 
complementary angle : 



sin 30°. 
cos 45°. 
tan 89°. 
cot 15°. 

sin 30° = cos 
cos 45° = sin 
tan 89° = cot 
cot 15° = tan 

esc 18° 10' 



esc 18° 10'. 
cos 37° 24'. 
cot 82° 19'. 
esc 54° 46'. 

(90° - 30°) = cos 60°. 
(90° - 45°) = sin 45°. 
(90° - 89°) = cot 1°. 
(90°- 15°) = tan 7 5°. 



sec (90° - 18° 10') 

= sec 71° 50'. 
cos 37° 24' = sin (90° - 37° 24') 

= sin 52° 36'. 
cot 82° 19 = tan (90° - 82° 19') 

= tan 7° 41'. 
esc 54° 46' = sec (90° - 54° 460 

= sec 35° 14'. 



2. Express as functions of an 
angle less than 45° : 

sin 60°. esc 69° 2'. 

cos 75°. cos 85° 39'. 

tan 57°. cot 89° 59'. 

cot 84°. esc 45° 1'. 

sin 60° = cos (90° - 60°) = cos 30°. 
cos 75° = sin (90° - 75°) =. sin 15°. 
tan 57° = cot (90° - 57°) = cot 33°. 
cot 84° == tan (90° - 84°) = tan 6°. 

esc 69° 2' = sec (90° - 69° 2') 

= sec 20° 58'. 
cos 85° 39' = sin (90° - 85° 39') 

= sin 4° 21'. 
cot 89° 59' = tan (90° - 89° 59') 

= tan 0° 1'. 
esc 45° 1' = sec (90° - 45° 1') 

= sec 44° 59'. 



teachers' edition. 



13 



3. Given tan 30° = i V3 ; find 

cot 60°. 

tan 30° = cot (90° - 30°) 

= cot 60°. 

.-. cot60° = i V3. 

4. Given tan A = cot A ; find A. 

tan A = cot (90° - A), 
90° - A = A, 
2 A = 90°. 
.-. A = 45°. 

5. Given cos A = sin 2 ^4 ; find A . 

cos .4 = sin (90°- ^4), 
90° -A =2 A, 
SA = 90°. 
.-. J. = 30°. 

6. Given sin A = cos 2 ^L ; find A. 

sin A = cos (90° - vl), 
90° _ ^4 =2 A, 
3A = 90°. 

.-. -4. =30°. 

7. Given cos ^4 = sin (45° — $A); 
find A. 

cos A =sin(90°- -4), 
90° -4 =45° -£4, 

180° -2 ^4 = 90° -.4. 

.-. ^4 = 90°. 

8. Given cot £ A — tan ^4; find A. 

tan ^4 = cot (90° -^4), 

\A - 90°- ,4, 



A = 180° -2^4, 
3^4 = 180°. 

.-. ^4 = 60°. 

9. Given tan (45° + A) — cot A 
find ^4. 

cot A = tan (90° - ^4), 
tan (90° - ^4) = tan (45° + A)< 
90° -A = 45° + ^4, 
2 A = 45°. 
.-. A = 22° 30'. 

10. Find A if 6in A = cos 4 A. 

sin A = cos (90° -^4), 
90° -A = 4^4, 

5^4 = 90°. 

.-. ^4 = 18°. 

11. Find A if cotyl = tan .8-4. 

cot A = tan (90° -A), 
$A = 90°-A, 
9 A =90°. 
.-. ^4 = 10°. 

12. Find J. if cot A — tan ?iyi. 

cot A = tan (90° - ^4), 
90°- A = n^l, 

90° = A (n + 1). 
90° 



^4 =• 



71 + 1 



EXEKCISE V, 

1. Prove Formulas [1], [2], [3], j 
using for the functions the line j 
values in the unit circle given in | 
Sect. Ill, p. 8. 

[1] sin 2 A + cos 2 A = 1. 
sin A 



m 



tan A = 



cos A 



Page 14. 

[3] sin A x esc A =1, 
cos ^4 x sec A = 1, 
tan J. x cot J. = 1. 
[1] ifP = sin .4, 

03f = cos .A. 
SiP 2 * OJ? 2 = OP 2 ; 
but OP 2 = 1. 



14 



PLANE TRIGONOMETRY. 



.-. MP 2 + OM 2 = 1. 
. sin 2 A -f cos 2 A = 1. • 
[2] MP = sin A, 

OM = cos A, 
AT= tan J.. 
A OJ. T and OlfP are similar. 
.-. TA:OA ::PM:OM. 
TA PM , 
(M ~ OM' 
OA =1. 

Pif 

sin .A 



Or, 
but 



TA = 



tan A = 



cos A 





p 






i 






^-^ 


\ 




tf 




y' 


T 






/ 


\ 








•i 




\ 


A 









J 


/ 



[3] 
and 



Or, 



B' 
Fig. 3. 

P3f = sin A, 
OS = esc A. 
In similar A OSB and P03f, 
OS: OB:: OP: PM. 
OS _ OP , 
OB ~ PJW ' 
but 05 = 1, 

and OP = 1. 

1 
PJf" 
.-. OS x PM = 1. 
.-. esc A x sin A. = 1. 
Again, cos A = OM, 
and sec A — OT. 



0S = 



Or, 



0T = 



In similar A OTA and 0P3f, 
OT: OA:: OP: OM. 
OT _ OP. 
0A~~ OM' 
but OA = 1, 

and 0P=1. 

1_ 

6m' 

.-. OTx 0M=1. 
.-. sec A x cos A. = 1. 
Also, tan A = AT, 
cot A = BS. 
In similar A SOB and TAO, 
BS:BO::AO:AT. 
BS _A0 . 
P0~ Zr' 
but PO = 1, 

and A0=1. 

AT 

.-. cot A x tan J. = 1. 



2 . Prove that 1 + tan 2 A - sec 2 A . 

tan A. = - , sec A = - • 

a 2 + 6 2 = c 2 . 
Divide all the terms by b 2 . 
a 2 6 2 _ c 2 
62 "^fc 2 "b 2 ' 



Or, 



a* 



j2 c 2 

Substitute for — and — 
b 2 b 2 

values tan 2 A and sec 2 A, 

tan 2 J. + 1 = sec 2 A. 



their 



3. Prove that 1 + cot 2 A = esc 2 A . 

b 

— — i 
a 

_ c 

~ a 

a? + 52 = C 2. 



cot A 
esc A 



TEACHERS' EDITKiV. 



15 



Divide all the terms by a 2 , 
a 2 b 2 c 2 

-> + - = -v 

a- a 2 a 2 

b 2 c 2 

Substitute for — and — their 
a 2 a 2 

values cot 2 ^4 and esc 2 ^4, 
1 -f- cot 2 A == esc 2 A. 



7. Prove that cos A esc A = cot A . 

4 b 
cos A = - , 
c 



CSC ^4. 



cot ^4 = 



4. Prove that cot^. = 



cot A = -• 
a 



cos ^4 
sin ^4 



sin A = 
cos ^4 = 
Substitute, 

.-. cot J. = 



a 

c 

b 

c 
b _b^a 
a~ c ' c 

cos A 



Substitute, - x - = 
c a a 

.-. cos A esc A = cot A. 

8. Prove that tan A cos A = sin A, 

a 

v 

b 
c 
a 
c 



sin A 



5 . Prove that sin A sec A = tan -4 . 

sin ^4 = - , 
c 

sec A = - , 
tan ^4 = - • 



Substitute, - x - = 



tan A 
cos A 
sin .4 



c b b 
.-. sin ^4 sec A = tan J.. 




.-. sin A sec J. cot J. = 1. 
10. Prove that 


6. Prove that sin A cot A - 

A a 

sin A = - , 
c 


=cosA. 


cos A esc A tan A = I. 

, b 
cos A = - , 
c 


. b 
cot A — - , 
a 




csc ^a = - , 

a 


cos ^1 = - • 

c 




tan ^4 = - • 


Substitute, - x - = - ■ 
c a c 




Substitute, - x - x - = 1. 
c a b 



.-. sin A cot A = cos A. 



^ i • (l b a 

Substitute, - x - = - • 
b c c 

.-. tan A cos A = sin A. 
9. Prove that sin J. sec ^1 cot J. — 1. 
sin A = - 1 



sec A • = - , 

6 



cot ^4 



a 



Substitute. - x - x - = 1. 
c b a 



.\cosAcscAtd,iiA — 1. 



16 PLANE TRIGONOMETRY. 

11. Prove that (1 - sin 2 ^.) tan 2 .4 = sin 2 ^i. 
From [1], 1 - sin 2 ^4 = cos 2 ^4. 

.-. (I - sin 2 A) tan 2 A = cos 2 .4 tan 2 ^4. 
But from Example 8, cos A tan A = sin A. 

.-. cos 2 ./! tan 2 yl = sin 2 ^t. 

12. Prove that Vl — cos 2 .4 cot A = Cos A. 



From [1], Vj — cos 2 ^. = sin ^4. 

.-. vl — cos 2 ^4 cot A = sin A cot A. 
But from Example 6, sin A cot A — cos A. 

.-. vl — cos 2 A cot A — cos A. 

13. Prove that (1 + tan 2 A) sin 2 A - tan 2 ^4. 

From Example 2, 1 + tan 2 .4 = sec 2 A. 

.-. (1 + tan 2 ^L) sin 2 A = sec 2 A sin' 2 A. 
But from Example 5, sec A sin A — tan A. 

.-. (1 + tan 2 Jl) sin 2 .4 = tan 2 A. 

14. Prove that (1 — sin 2 ^t) esc 2 A — cot 2 ^4. 

From [1], 1 — sin 2 ^4 = cos 2 ^i. 

.-. (1 — sin 2 A) esc 2 A = esc 2 A cos 2 A. 
But from Example 7, esc A cos A — cot A. 

.-. (1 - sin 2 ^4) esc 2 A = cot 2 A. 

15. Prove that tan 2 .4 cos 2 A -f cos 2 A — 1. 

From Example 8, tan A cos A = sin A. 

.-. tan 2 A cos 2 A = sin 2 ^4. 
tan 2 A cos 2 A -f cos 2 -4 = sin 2 ^4 + cos 2 A. 
From [1], sin 2 A + cos 2 A = 1. 

.-. tan 2 ^4 cos 2 ^4 + cos 2 J. = 1. 

16. Prove that (sin 2 A — cos 2 J.) 2 = 1-4 sin 2 A cos 2 ^4. 
From [1], sin 2 A 4- cos 2 A = 1. 

.-. (sin 2 .4 4- cos 2 ^.) 2 = 1. 
But from Algebra, (sin 2 A — cos 2 ^4) 2 = (sin 2 A 4- cos 2 ^4) 2 

— 4 sin 2 ^4 cos 2 A. 
.-. (sin 2 A — cos 2 J.) 2 = 1—4 sin 2 A cos 2 A. 

17. Prove that (1 - tan 2 ^.) 2 = sec 4 A - 4 tan 2 .4. 
From Example 2, 1 -f- tan 2 .4 == sec 2 ^.. 

.-. (1 + tan 2 ^4) 2 =:sec 4 ^4. 



TEAUHEKS EDITION. 1* 

But from Algebra, (1 - tan* A) 2 = (1 -f tan 2 A) 2 - 4 tan 2 A. 
.-. (1 - tan 2 A) 2 = sec 4 A - 4 tan 2 A. 

_ _ , sin A cos A 

18. Prove that 1 — = sec A esc A . 

cos A sin A 

sin A cos^l _ sin 2 ^4 + cos 2 ^l 
cos A sin A cos ^4 sin A 

But from [1], sin 2 A + cos 2 A = 1. 

And from [3], — — - = sec A, 

cos A 

and = esc A. 

sin A 

sin A cos A 

.-. 1 = sec A esc A. 

cos A sin A 

19. Prove that sin 4 A — cos* A — sin 2 A — cos 2 A. 

sin 4 -4 — cos 4 ^4 = (sin 2 A -f cos 2 A) (sin 2 A — cos 2 ^4). 
From [1], sin 2 A + cos 2 ^4 = 1. 

.-. sin 4 A — cos 4 A = sin 2 -4 — cos 2 A. 

20. Prove that sec A — cos A = sin A tan ^4. 

1 



From [3], sec A 



cos A 



8 e e A-cosA=J-- C osA 

cos A 

1 — cos 2 A 



cos A 

From [1], 1 — cos 2 A = sin 2 J.. 

sin 2 ^4 

.-. sec ^4 — cos .A = 

cosA 

m-, Sill -O. . 

Also, from [2], = tan A. 

cos A 

.-. sec A — cos A = sin A tan ^4. 



21. Prove that esc A — sin A = cos ^4 cot J.. 

1 



From [3], 





CSC ^L = 

sin A 


. csc A - 




sin ^4 




_ 1 - sin 2 A 



sin A 



18 



PLANE TRIGONOMETRY. 



From [1], 



Also, from [2], 



1 — sin 2 A = cos 2 A. 
cos 2 A 



.-. esc A — sin A •. 



cos A 



sin .4 
cot J.. 



sin^L 
esc A — sin A = cos A cot A. 



22. Prove that 



cos A 



1 + sin A 



1 — sin A cos A 
Clearing of fractions, this becomes, 

cos 2 A = 1 — sin 2 .4, 
which is correct by [1]. 

cos A 1 + sin A 



1 — sin A 



cos A 



Exercise VI. Page 16. 



1. Find the values of the other 
functions when sin A = ■$■■ }. 



sin 2 A + cos 2 A 
cos 2 A 

cos A 



1, 

1 - 






25^ 
169' 



,\ cos J. = 

13 



tan ^1 



sin J. 



12 
J' 



cos A 
cot .A is reciprocal of tan A. 

.-. cot A = — 
12 

sec A is reciprocal of cos A. 

.-. sec A = — 
5 

esc A is reciprocal of sin A. 

13 



.-. esc A — ■ 



12 



2. Find the values of the other 
functions when sin A =0.8. 

sin 2 A + cos 2 A = 1, 

cos 2 J. = 1 -(0.8) 2 , 



cos A 


= Vl -0.64. 


.-. cos A 


= 0.6. 




tan A 


sin A 
cos A 


0.8 
0.6 


.-. tan^i 


= 1.3333. 




cot .4 


_0.6 
~0.8* 




.-. cot A 


= 0.75. 




sec A 


1 
~(X6* 




.-. sec A 


= 1.6667. 




csc .4 


1 

~0.8* 




.-. esc A 


= 1.25. 





3. Find the values of the other 
functions when cos A = f f . 

sin 2 A -f cos 2 A — X) 



TEACHERS' EDITION. 



19 



U = ^Z 




tan .4 



cot .4 = 



sec A = 



esc A 



4. Find the values of the other 
functions when cos A = 0.28. 
sin 2 ^4 + cos 2 A = 1, 
sin ^4 

tan A = 

cot A = 

sec A = 

esc A = 

s'mA 0.96 



5. Find the values of the other 
functions when tan A = f . 



= Vl 


- (0.28) 2 




= Vo. 


9216 = 0.96. 


sin A 
cos A 


_0.96 _ 
~~ 0.28 ~ 


3.4286. 


1 


1 


— 291 


tan A 


3.4286 




1 
cos A 


1 

"" 0.28 ~ 


3.5714. 


1 


1 


1.0417. 



cos 


A 


= 


sin A 




3 


tan ^4 


5 


sec 


A 


= 


1 




5 


cos A 


3 








1 




5 


CSC 


A 


— 




— 




sin A 


4 



6. Find the values of the other 
functions when cot A = 1. 



tan ^L 


9 * 


cot A 


3 
~4* 


tan ^4 


sin A 
cos A 


4 


sin A 


3 


Vl — sin 2 J. 



3 sin ^4 =4 Vl - sin 2 X 
9 sin 2 J. = 16- 16 sin 2 A, 
5 sin ^4 = 4. 

.-. sin A — -. 



cot ^4 


= 1. 




.-. tan J. 


= 1. 




tan A 


sin ^4 
cos ^4 




1 


sin ^4 




vrr 


sin 2 ^4 


sin ^4 


= Vl- 


sin 2 ^4, 


sin 2 A 


= 1 - sin 2 ^4, 


2 sin 2 J. 


= 1, 




sin 2 ^4 


_ 1 
~2* 




.-. sin A 


= Vj = 


fV2. 


cos ^4 


sin J. 
tan J. 


= iV£ 


sec A 


1 
cos -4 


i 

-J-V2 




1 


1 



sin 4 



-V2 



:V£. 



-VS. 



7. Find the values of the other 
functions when cot A = 0.5. 

1 1 



tan A — 
tan A = 



cot A 
sin A 



0.5 

:2. 



cos ^4 
2 cos A = sin A. 

4 cos 2 ^4 — sin 2 A — 0. 

cos 2 A -f sin 2 ^4 = 1. 

Add, 5 cos 2 ^4 = 1. 



20 



PLANE TRIGONOMKTKY. 



COS 



A=yjl = 0.45. 



4 cos 2 A -f 4 sin 2 ^4 = 4 

4 cos 2 A — sin 2 A — 

5 sin 2 A —\ 

sin A = a/- = 0.90. 



sec A = = 2.22. 

cos A 

esc A - =1.11. 

sin A 



8. Find the values of the other 
functions when sec A = 2. 
1 _ 1 
~2' 



cos A. = 



sec A 
sin A = Vl - 



V> 



-"-i- 



.-. sin A — i V3. 

sin ^4 
tan ^4 == 

cot A = 



cos v4 

1 
tan J. 



esc ^4 = 



cos 2 A 
\4 

I 

1 

V3 = 
= 1^3. 



4.V3. 



sin J. 

9. Find the values of the other 
functions when esc A = V2. 

sin A = — = 1 V2. 
V2 " 



i A = Vi - (i V2) 2 = Vi - 1 



tan J. 



cot A 



= VT = 1 V2. 
& V2 



iV5" 



sec A = 



V2 



= V5. 



10. Find the values of the other 
functions when sin A = m. 
cos A 

m 



tan A 

cot A = 
sec A = 
esc ^4 = 

11. 



Vl - sin 2 A = Vl 
sin J. _ 



m z . 



cos ^4 

?n v 1 — m 2 



m L 



m 2 



— ^— = - Vl - m 2 . 
tan A m 

1 1 



cos A 

1 
sin yl 



Vl - m 2 
1 



Find the values of the other 
functions when sin A = 



1 + m 2 



cos A = Vl 
, cos .4 =V/i 



sin 2 J.. 



4 m 2 



1 + 2 ??i 2 + ??i 4 



=Vr 



2 ?n 2 + m 4 



-f 2 m 2 + m 4 



1 



tan ^4 = 
cot A = 
sec A = 
esc A = 



1 + m 2 
sin ^4 



cos A 

1 

tan ^4 

1 
cosA 

1 
sin A 



2 m 
1 -m 2 ' 
1 -m 2 

2m 
1 + m 2 
1 - m 2 ' 
1 + m 2 

2 m 



12. Find the values of the other 
functions when cos A = 



m 2 + n 2 



sin A = Vl — cos 2 A. 



4 m 2 w 2 



m 4 -f 2 ?n 2 n 2 + 



3 



TEACHERS EDITIOX. 



21 



-V 



tan A = 



fm 4 - 2 ra 2 n 2 + n 4 
w 4 + 2 ?n 2 n 2 + ?i 4 
m 2 — n 2 

' m 2 + n 2 ' 
sin ^4 



cot ^4 = 
sec ^4 

esc A 



cos ^4 
1 



2 run 
2mn 



tan A m 2 — n 2 
1 _ m 2 + n 2 

cos ^1 2 ?n?i 
1 m 2 + n 2 

^n 2 ' 



sin ^4 ?n 2 

13. Given tan 45° 
other functions of 45° 

sin 45° 



1 ; find the 



cos 45° 
sin 45° 



tan 45° 



= 1. 



cos 45° 
sin 2 45° + cos 2 45° = 1. 
By (1), sin 45° = cos 45°. 
By (2), cos 2 45° + cos 2 45° 
2 cos 2 45° = 1, 

cos 2 45° =-, 



(1) 
(2) 



cos 45° = 



5 o = ^± = iV2. 



sin 45° = i V2. 

i 
cot 45° : 



sec 45° : 



tan 45° 
1 



_ 1 
~ 1 ~ 

V5. 



esc 45° = 



i^2 

J-=V2. 
/5 



* v 

14. Given sin 30° = $ ; find the 
other functions of 30°. 

sin 2 30° + cos 2 30° = 1. 



tan 30° = -JL_ = l V3. 

i V3 



cot 30° = = V3. 

1V3 



sec 30° : 



esc 30° 



j V5 



= 4 VS. 



15. Given esc 60° = f V3 ; find 
the other function of 60°. 

1 



sin 60° 



esc 60 D 
1 



fV3 



-1 V3. 



cos 60° = Vl -sin 2 60° 

= Vi _ (i V3) 2 

\ 4 2 



= Al 



I V3 /- 
tan 60° = 2_ — = V3. 



cot 60° = 



Vs 



sec 60° = - = 2. 
1 



=w§. 



16. Given tan 15° = 2 - V3; find 

i he other functions of 15°. 

sin 15° n rz 

tan lo° = = 2 - V3. 

cos 15° 

sih 2 15° + cos 2 15° = 1. 
sin 15° = (2 - V3)cosl5°. 
[(2 - V3) cos 15 ] 2 + cos 2 15° = 1, 
(4 - 4 V3 + 3) cos 2 15° + cos 2 15° 

= 1, 
(8-4 V3)cos 2 15° = l. 

1 2 +V3 



cos 2 15° 



4 (2 - V3) 



22 



PLANE TRIGONOMETRY. 



cos 15° = \/ 2+ 8 = J V2 + V3. 

sin 2 15° = 1 -cos 2 15°. 

sin215 o = l- 2 -±^ = 2 -^?, 



sin 15° = \ V2 - V3. 



cot 15° = 



1 



1 



sec 15° = 



tan 15° 2 — V3 
= 2 + V3. 
1 



esc 15° = 



1 V2 + V3 

2 (2 - V3) V2 -f V3. 
1 

1 V2 - V3 

: 2 (2 + V3) V2 - V3. 



17. Given cot 22° 30' = V2 + 1 ; 
find the other functions of 22° 30'. 

1 1 



" 2 cot 221° V2 + 1 




= V2-1. 




Sin22 " = tan 22io, 
cos 221° 


a) 


cos 2 22J + sin 2 22i = 1. 


(2) 


From (1), 




cos 22J° tan 22^° = sin 22i°. 




Square, 





cos 2 22 J° tan 2 221° = sin 2 22^°. 
From (2), 

cos 2 22i° = -sin 2 22i°+ 1 



Add, 
cos 2 22i tan 2 22J° + cos 2 22i° = 1. 
cos 2 22i°(tan 2 22J + 1) = 1, 
cos 2 22J°(4-2 V2) = l, 
cos22-J-° V4-2 V2 = 1. 
.-. cos22',° = - 



V4 - 2 V2 



4 



4 + 2 V5 



1 V2 + V2. 



sin 22 



l° = \l- 



2+V2 






V2 



sec 221° : 



: £ V2 - V2. 

1 



csc 22.^° 



W2+V2 

= (2 - V2) V2 + V2. 
1 



1 V2 - V2 

= (2 + V2) V2 - V2. 

18. Given sin 0° = ; find the 
other functions of 0°. 



cos 0° = Vl - sin 2 0° 

=rl. 

, AO sin 0° _ 
tan 0° = — = - = 0. 



cos 0° 1 



cot 0° 



1 



1 



tan 0° 



= - = 00. 



sec 0° - - 



cos 0° 1 



csc 0° : 



1 



1 



sin 0° 



19. Given sin 90° = 1 ; find the 
other functions of 90°. 
sin 99° = 1. 



cos 



90° = Vl - sin 2 90° = 0. 

+ OAO sin 90° 1 

tan 90° = = - = 00. 

cos 90° 



TKA< HKi;>' EDITION. 



23 



cot 90 = 
sec 90° = 
esc 90° = 



-L- - = i = o. 

tan 90 x 



1 _ 1_ 
cos 90- " ~ 

1 



cot .1 
sec -4 = 
esc A 



cos ^4 



Vl — cos 2 ^4 

1 



cos A 



20. Given tan 90° = x ; find the 
other functions of 90°. 

tan 90 - = x. 



tan 90° x 






cot 90- = 

^^ = cot 90^ = 0. 
sin 90° 

. cos 90 : = 0. 

sin-90 : - 008*90° = 1. 

. sin-90 r = 1, 
sin 90° = 1. 

sec 90 = - = x . 


esc 90 : = 1. 

21. Express the values of all the 
other functions in terms of sin A. 

cos A = 

tan A = 

cot .4 = 
sec A = 
esc A = 



\ 1 - cos- A 

23. Express the values of all the 
other functions in terms of tan A. 
1 



cot A = 

a 
b 



tan A 
■ tan A. 



a = h tan A. 

a- = b 2 tan- A. 
a- - 6- tan 2 .4 =0 
gjhjg =1 

6-(l ~ tan-\4) = 1 

1 



p = 



1 + tan* A 



= cos 2 . 4, 



cos .1 = 



Vrn 



an- A 



vi 


— sin 2 

sin A 


A. 


\ i 


- sin"- 


A 


\ i 


— sin 2 


A 


sin A 

1 




M 

1 


— sin 2 


A 



sin .4 



sin A = % 1 — cos- A 



1 



1 — tan- J. 
tan A 



sec A = 
esc .4 = 



Vl + tan- .4 

1 



cos .4 

1 
sin ^4 



\ 1 4-tan*4. 



V 1 + tan-.4 
tan A 



22. 

other 



24. Express the values of all the 
Express the values of all the other f unctions in terms of cot A . 

functions in terms of cos .4. 

1 

sin A = VI _ cos 2 ^4. 



tan A = 



VI - cos- .4 



eos.4 



cot .4 
sin J. 
cos ~4 



= tan J. . 
tan A. 



24 



PLANE TRIGONOMETRY. 



Let 



x = sin -A, y = cos A. 
x 1 



y cot A 
x cot ^4 = ?/, 
x* 2 cot 2 ^4 = y 2 . 

x 2 cot 2 A -y 2 = 



x 2 



+i! 



x 2 (l +cot 2 ^) = l 

1_ 

1 + cot 2 ^l' 



x 2 = 



1 



sin A — — ;-- .:,—--■ 
Vl + cot 2 -A 

cos J. = Vl — sin 2 A 



4 



= \l 



1 4- cot 2 .4 
/l 4- cot 2 J. - 1 



=v : 



1 + cot 2 .4 
cot ^4 



sec .A = 



Vl + cot 2 .4 
1 Vl + cot 2 J. 



esc A 



cos ^4 cot A 

1 



sin J. 



: Vl +cot 2 ^4. 



25. Given 2 sin A = cos ^4; find 
sin A and cos ^4. 

sin 2 .4 -f cos 2 ^4 — 1. 
sin 2 A + 4 sin 2 ^4 = 1. 
5 sin 2 A = 1, 
1 



sin 2 ^4 = 



sin A = 



■ = i Vs. 



cos 



.A = I Vs. 



26. Given 4 sin ^4 = tan ^4 ; find 
sin A and tan ^4. 

sin A 



tan ^4 = 



cos A 



it tan ^4 = 


4 sin A. 




.-. sin A == 


sin A 




cos J. 




4 sin A x 


cos A — 


sin A. 


.-. cos ^4 = 


sin A 
4 sin ^4 


1 
"I' 


sin 2 ^4 4- 


cos 2 ^4 = 


1. 


.-. sin ^4 = 


xM 


-w 


= 


i Vl5. 




tan A = 


sin ^4 

cos A 




= 


iVl5_ 

■i 


Vl5. 



27. If sin J. : cos ^4 = 9 : 40, find 
sin J. and cos A. 

40 sin ^4 = 9 cos A. 

Square, 

1600 sin 2 A = 81 cos 2 A. 

1600 sin 2 A - 81 cos 2 A = 0. 
But sin 2 A 4- cos 2 A = 1. 

Multiply by 81 and add, 

1681 sin 2 A = 81. 

.-. 41 sin ^4=9. 

sin ^4 = 



41 
sin 2 A 4- cos 2 v4 == 1. 

cos A = Vl — sin 2 J.. 

L / 9 \* 40 

.-. cos A =-\/l— ( — )= — . 
\ \41/ 41 

28. Transform the quantity 
tan 2 A 4- cot 2 ^4 - sin 2 J. - cos 2 A 
into a form containing only cosv4. 

. sin 2 A 1 — cos 2 ^4 

tan 2 A = = 

cos 2 A cos 2 ^4 

. A cos 2 u4 cos 2 A 

cot 2 A = = - • 

sin 2 A 1 — cos 2 A 



TEACHEKS EDITION. 



25 






1 — COS 2 ^4 cos 2 ^4 

cos 2 A 



1 — cos 2 ^4 
— 14- cos 2 A — cos 2 A 
1 — 2cos 2 -4 4-2cos 4 ^4 — cos 2 ^. + cosM. 



1 



cos 2 ^. — cos 4 ^4 
■3 cos 2 A + 3 cos 4 ^4 



cos 2 ^4 — cos 4 A 



30. Prove that 
tan A 4- cot A = sec A x esc A . 



29. Prove that 
sin A 4- cos A = (1 4- tan A) cos A. 

Psin A 
= tan A . 
cos A 

sin A = tan ^4 cos A. 

sin J. + cos A— tan ^4 cos ^4 4- cos A 

— (1 4- tan ^4) cos ^4. 

Exercise YII 

1. Solve the equation 
2 cos x = sec x. 
1 



tan A — 

cot ^4 = 

tan A + cot .4 = 



sin A 
cos ^4 
cos A 
sin ^4 
sin ^4 



■ + ■ 



cos A 



cos A sin ^4 
sin 2 ^4 + cos 2 A 



But sin 2 ^. 
. tan A + cot A = 



cos ^4 sin ^4 

cos 2 ^4 = 1. 
1 



2 cos x = 

, 2 cos 2 x = 



cosx 
1. 

cos X = V^-. 
.-. x = 45°. 

2. Solve the equation 

4 sin x = esc x. 

( . 1 

4 sm x = 

sin x 

.-. 4sin 2 x = 1. 

sin x = -J-. 

.-. x = 30°. 

3. Solve the equation 

tan x = 2 sin x. 
sinx 



: 2 sin x. 



cosx 

sin x = 2 sin x cos x. 
(1 — 2 cosx) sinx =0. 
.-. sinx = 0. (1) 



cos A sin A 
= sec A x esc A. 

Page 18. 

.-. x = 0. 

1 — 2 cosx = 0. (2) 

cos x = £. 
.-. x = 60°. 
.-. x = 0° or 60°. 

4. Solve the equation 

sec x = V2 tan x. 

cos X COS X 

1 = V£ sin x. 
1 

v-T 

45°. 



sm x = 



.-. x : 

Solve the equation 
sin 2 x = 3 cos 2 x. 
sin 2 x _ 

— o. 



cos 2 x 
tan 2 x = 3. 
tan x = V3. 
.-. x = 60°. 



26 



PLANE TRIGONOMETRY. 



6. Solve the equation 

2 sin 2 x -f cos 2 x = §. 

sin 2 x 4 (sin 2 x -f cos 2 x) = f . 

sin 2 x 4- 1 = |. 

sin 2 x = J. 

1 

sin x = -— • 
V2 

.-. x = 45°. 

7. Solve the equation 
3tan 2 x — sec 2 x = 1. 

3 tan 2 x - (tan 2 x + 1) = 1. 

2tan 2 x- 1 = 1. 

2tan 2 x = 2. 

tanx = 1. 

.-. x = 45°. 

8. Solve the equation 

tan x 4- cot x = 2. 
1 



tan x -f- 



• = 2. 



tanx 

tan 2 x - 2 tan x -f 1 = 0. 
tan x — 1 = 0. 
.-. tan x = 1. 
.-. x = 45°. 

9. Solve the equation 

sin 2 x — cos x = I. 

(1 - cos 2 x) — cosx = i. 

cos 2 x 4 cos x = f. 

cos 2 x + cosx 4 i = 1. 

cosx + | = 1. 

cos x — \. 

.-. x = 60°. 

10. Solve the equation 

tan 2 x — secx= 1. 

(sec 2 x - 1) -secx = 1. 

sec 2 x — secx = 2. 

sec 2 x -- secx -f i = f- 

sec x - i = ± f . 
sec x = 2. 
.-. x = 60°. 



11. Solve the equation 
sin x + V3 cos x = 2. 

V3 cos x = 2 — sin x. 
3cos 2 x = (2 -sinx) 2 . 
3 (1 - sin 2 x) = 4 - 4 sin x + sin 2 x 
4 sin 2 x — 4 sin x + 1 =0. 
2 sin x — 1 = 0. 
sin x = -J-. 
.-. x = 30°. 

12. Solve the equation 

tan 2 x + csc 2 x = 3. 

tan 2 x + (1 + cot 2 x) = 3. 

tan 2 x-2 -fcot 2 x = 0. 

tan x — cot x = 0. 

1 

tan x = 0. 

tanx 

tan 2 x = 1. 

tan x = 1. 

.-. x = 45°. 

13. Solve the equation 

2 cos x 4- sec x = 3. 
1 



2 cos x -f 



3. 



cosx 
2cos 2 x 4 1 = 3 cosx. 
2 cos 2 x — 3 cos x 4 1 = 0. 
(2 cosx - l)(cosx - 1) = 0. 

cos x = 1 or |. 
.-. x = 0° or 60°. 






14. Solve the equation 
cos 2 x — sin 2 x = sin x. 

(1 - sin 2 x) - sin 2 x = sin x. 
2 sin 2 x + sin x — 1 = 0. 
(2sinx - l)(sinx 4 1) = 0. 

sin x = — 1 or J. 
.-. x = 30°. 

15. Solve the equation 

2 sin x 4 cot x = 1 4 2 cos x. 



TEACHERS' EDITION'. 



27 



2 sin x H — - = 1+2 cos x. 

sin x 

2sin 2 x + cosx 

= sin x 4 -2 cos x sin x. 

2 sin 2 x — sin x 

= 2 cos x sin x — cos x. 

sinx(2 sinx — 1) 

= cosx(2 sin x — 1). 

(sin x — cos x) (2 sin x — 1) = 0. 

.*. sinx = cosx. 

tanx = 1. 

/. x = 45°. 

sin x = i. 

.-. x = 30°. 

Hence x = 30° or 45°. 

16. Solve the equation 

sin 2 x 4- tan 2 x = 3 cos 2 x. 

. sin 2 x * 2 

sm 2 x H = 3 cos 2 x. 

cos 2 x 



1 — cos 2 x 
1 — cos 2 x -} — = 3 cos 2 x. 



— 4 cos 2 x + 



cos^x 
1 



= 0. 



COS 2 X 

4 cos 4 x = 1. 
cos X = V-J-. 
/. x = 45°. 

17. Solve the equation 

tan x 4- 2 cot x = f esc x. 

sin x cos x _ 5 
cosx sin x 2 sinx 
sin 2 x 4- 2 cos 2 x = f cos x. 
1 — cos 2 x 4- 2 cos 2 x = | cos x. 
cos 2 x — | cos x + 1 = 0. 
(cosx — 2) (cosx — \) = 0. 

cos x = 2 or i. 
.-. x = 60°. 



Exercise VIII. Page 24. 



1. In Case II give another way 
of finding c, after b has been found. 



cos^l = -• 
c 



b = ccos^l. 
b 



.'. C : 



cos A 



2. In Case III give another way 
of finding c, after a has been found. 



sin .A = 

c sin A = 

.-. c = 



a 



sin J. 



3. In Case IV give another way 
of finding b, after the angles have 
been found. 



cos A = - • 
c 

.-. b = c cos A. 

4. In Case V give another way 
of finding c, after the angles have 
been found. 

sini = - • 



c sin A — a. 






a 



sin A 

5. Given B and c ; find A, a, b. 
A = 90° - 5. 

cos B = - • 
c 

.-. a = c cos B. 

c 
.-. 6 = c sin JB. 



28 



PLANE TRIGONOMETRY. 



6. Given B and b ; find A, a, c. 
A = 90° - B. 

cot £ = -. 

6 

.-. a = 6 cot JB. 

sin J5 = - . 
c 

b 

.*. c 



sinB 
7. Given B and a ; find A, 6, c. 



cot B . 

.'.6: 

COS jB: 

.'. C : 



A = 90° - JS. 
a 



a 



cot 5 

a 
c 



= atanJ5. 



a 



cos I? 

8. Given b and c ; find A, JS, a. 
b 



cos ^4 



B = 90° - 4. 



a = V C 2 - 62 



= V(c + 6)(c-&). 

9. Given a = 3, 6 = 4; required 
A = 36° 52', 5 = 53° 8', c = 5. 

tan .A = - = - = 0.7500. 
6 4 



.-. A = 36° 

i?^=90° 

= 53° 


52'. 

-J. 

8'. 


c = Va s 
= 5. 


+ & 2 



10. Given a = 7, c = 13 ; required 
4 = 32° 35', 5 = 57° 25', b = 10.954. 

sin A = - = — = 0.5385. 
c 13 



,\ J. = 32° 35'. 
B = 90°-A 
= 57° 25'. 

b = V(c - a) (c + a) 



= V120 
= 10.954. 

11. Given a = 5.3, JL = 12° 17' ; 
required £ = 77° 43', b = 24.342, 
c = 24.918. 

B = 90° - A 
= 77° 43'. 

- = cot .4. 
a 

.-. b — a Got A 

= 5.3 x 4.5928 

= 24.342. 

a . . 

- = sin A. 



.*. c = 



a 



sin A 
5.3 
~ 0.2127 
= 24.918. 

12. Given a = 10.4, B = 43° 
required A = 46° 42', 6 = 9.800, 
14.290. 

A = 90° - B 
= 46° 42'. 

- = tan B. 
a 

.-.b — a tan B 

= 10.4 x 0.9424 

= 9.800. 

a 

- — cos B. 



cosB 
10.4 

"0.7278 
= 14.290, 



TEACHERS 7 EDITION. 



29 



13. Given c = 26, A = 37° 42' ; 
required B = 52° 18', a = 15.900, 
b = 20.572 

jB = 90° - A 
= 52° 18'. 

- = sm A. 
c 

.-. a — c sin A 

= 26 x 0.6115 

= 15.900. 

b , 

- = cos A. 
c 

.-.b = c cos A 

= 26 x 0.7912 

= 20.572. 

14. Given c = 140, B = 24° 12' ; 

required ^L = 65° 48', a = 127.694. 
6 = 57.386. 

A = 90° - 7? 

= 65° 48'. 

- = cos B. 
c 

.-. a = ccos J5 

= 140 x 0.9121 

= 127.694. 

- = sm B. 
c 

.-. 6 = csin B 

= 140 x 0.4099 

= 57.386. 

15. Given b = 19, c = 23 ; required 
.4 = 34° 18', B = 55° 42', a = 12.961. 

b 19 

cos A = - = — = 0.8261. 

c 23 

.-. A = 34° 18'. 

.-. B = 90° - yi 

= 55° 42'. 



16. Given b = 98, c = 135.2 ; re- 
quired ^L=43°33 / , J5 = 46°27 / . a 
= 93.139. 

„ 6 98 

cos A = - = 

c 135.2 

= 0.7248. 
.-. A = 43° 33'. 
.-. B = 90° - A 

= 46° 27'. 



: COS .4. 



a = V(c - b)(c + 6) 

= Vl68 
= 12.961. 



a = V(c - b) (c + 6) 
= V8675.04 
= 93.139. 

17. Given 6 = 42.4, A = 32° 14'; 
required B = 57° 46', a = 26.733, 
c = 50.124. 

B = 90° - A 
= 57° 46'. 

- = tan A . 
b 

.-. a = b tanyl 

= 42. 4 x 0.6305 

= 26.733. 

b 

c 
_ b 

cos A 
42.4 
~ 0.8459 
= 50.124. 

18. Given b = 200, B = 46° 11' ; 
required A = 43° 49', a = 191.900, 
c = 277.160. 

A = 90° - B 
= 43° 49'. 

- = cotB. 
b 

.-. a = b cot 5 

= 200 x 0.9595 

= 191.900. 

- = sin B. 

c 



30 



PLANE TRIGONOMETRY. 





' 


sin B 










200 










~~ 0.7216 










= 277.160. 






19. 


Given 


a = 95, b = 


:37; 


re- 


quired 


A = 


68° 43', B = 


21° 


17', 


c = 101.951. 










tan 


a_95_ 


2.5676. 







A: 


= 68° 43'. 










B 


= 90° - A 










c 


= 21° 17'. 








= Va 2 + 6 2 






= V10394 










= 101.951. 






20. 


Given 


a 


= 6, c = 103 ; 


re- 


quired 


A = 


3° 


21', B = 


86° 


39', 


b = 102.825. 












sir 


iA 


_a_ 6 
" c~~103~ 


= 0.0583. 






A 


= 3° 21'. 








•" 


B 


= 90° -A 
= 86° 39". 







.'. c = 



cosB 
3.12 
~ 0.9960 
= 3.133. 

22. Given -a = 17, c = 18; re- 
quired A = 70° 48', £ = 19° 12', - 
b = 5.916. 



tan I B -. 



4 
-4 



a 



*c + a 

if 35 

= 0.1690. 
•. |J5 = 9°36'. 
.-. 5 = 19° 12'. 

.-. ^4 = 90° - £ 
= 70° 48'. 



= V^ 2 ~-0 2 



= V10573 
= 102.825. 

21. Given a = 3.12, J5 = 5° 8'; 
required A = 84° 52', 6 = 0.280, 
c = 3.133. 

A = 90° - £ 
= 84° 52'. 
6 



a 



: tan B. 



.-. b = a tan B 
= 3.12 x 0. 
= 0.280. 

- = cos B. 

c 



b = V(c — a) (c + a) 
= V35 
= 5.916. 

23. Given c = 57, A = 38° 29'; 
required B = 51° 31', a = 35.471, 
b = 44.620. 

B = 90° -A 
= 51° 31'. 

a . . 

- = sm A. 
c 

.\ a = c sin J. 

= 57 X 0.6223 

= 35.471. 

- = cos A. 
c 

.\b = c cos vi 

= 57 x 0.7828 

= 44.620. 



24. Given a + c = 18, 6 = 12 ; re 
quired A = 22° 37', B = 67° 23 
a = 5, c = 13. 



1 



TEACHERS' EDITION. 



31 



c 2 - a 2 = b 2 . 

(c + a) (c - a) = b 2 . 

18 (c - a) = 144. 

c — a = 8. 

.-. c = 13. 

.-. a = 5. 



sin A = a = — = 0.3846. 
c 13 

.-. A = 22° 37'. 

.-. E = 90° - A 

= 67° 23'. 



25. Given a + b = 9, c = 8 ; re- 
quired A - 82° 18', B = 7° 42', a = 
7.928, b = 1.072. 



a 2 + ft 2 = c 2 = 04 
a 2 + 2 afr + 6 2 = 9 2 = 81 

2ab= 17 
a 2 - 2 a6 + b 2 = 64_- 17 = 47. 
.-. a -b= V47 = 6.856. 
But a + 6 = 9. 

.-. a = 7.928 
and 6 = 1.072. 



tan i B ■ 



■ p" 



0.072 



/ 15.928 
= 0.0672. 
.-. \ B = 3° 51'. 
.-. J5 = 7° 42'. 
.-. A = 90° -2? = 82° 18'. 



Exercise IX. Page 28. 



1. Given a = 6, c = 12 ; required 
A = 30°, B = 60°, b = 10.392. 

. " a -1 

sin A = - = -• 

c 2 

.-. A = 30°. 

£ = (90°- A) =60°. 

cos A = -• 
c 

.-. 6 = ccos A. 

log cos A = 9.93753 

log 12 = 1.07918 

log 6 =1.01671 

b = 10.392. 

2. Given A = 60°, 6=4; required 
£ = 30°, c = 8, a =6.9282. 



cos A 



B = 90° - A = 30°. 

6 



log 6 = 0.60206 
colog cos A = 0.30103 
logc =0.90309 
c = 8. 

c 2 = a 2 + b 2 . 
... C 2 _ 52 = a 2 _ 4 8 . 

log 48 = loga 2 = 1.68124. 
.-. log a = 0.84062. 
a = 6.9282. 

3. Given A = 30°, a = 3 ; required 

£ = 60, c = 6, 6 = 5.1961. 

B = (90° - A) = 60°. 

. . a 
sin A = - • 

c 

_ a 



cos A 



sin A 

log a = 0.47712 

colog sin A = 0.30103 

logc = 0.77815 

c = 6. 

c 2 = a 2 + b 2 . 



32 



PLANE TRIGONOMETRY. 



.-. c 2 - a 2 = 6 2 = 27. 

log 27 = log6 2 = 1.43136. 
.-. log 6 = 0.71568. 
6 = 5.1961. 

4. Given a = 4, 6 = 4; required 
A = B = 45°, c = 5.6568. 

Since a and 6 each = 4, the A is 
isosceles and Z A = Z B. 

.-. A = i of 90° = 45°, 
and B = i of 90° = 45°. 
c 2 = a 2 + 6 2 = 32°. 
log 32 = log c 2 = 1.50515. 
.-.logc = 0.75257. 
c = 5.6568. 

5. Given a = 2, c = 2.82843 ; re- 
quired A = B = 45°, 6 = 2. 

6 = Vc 2 — a 2 



= V(c -f a)(c - a), 
log 6 2 = log(c + a) 4- log(c — a). 
log(c + a) = 0.68381 
log(c - a) = 9.91826 - 10 
log 6 2 = 0.60207 
log 6 = 0.30103. 
6 = 2. 
.-. the A is an isosceles rt A. 
.-. A = B = 45°. 

6. Given c = 627, A = 23° 30 / ; 
required 5 = 66° 30 r , a = 250.02, 
6 = 575.0. 

B = (90° - A) = 66° 30^ 

a = c sin ^4. 

log a = log c -f log sin A. 

logc =2.79727 

log sin A = 9.60070 

log a = 2.39797 

a = 250.02. 

6 = c cos A. 



log 6 = log c + log cos A. 
logc =2.79727 
log cos A = 9.96240 
log 6 =2.75967 

6 = 575. 

7. Given c = 2280, A = 28° 5'; 
required J5 = 61° 55', a = 1073.3, 
6 = 2011.5. 

B = (90° - A) = 61° 55 r . 
a = c sin J.. 
log a = log c -f log sin A. 
logc =3.35793 
log sin A = 9.67280 
log a = 3.03073 
a = 1073.3. 
6 = c cos .A. 
log 6 = log c -h log cos A. 
log c = 3.35793 
log cos A = 9.94560 



log 6 


= 3.30353 




6 


= 2011.5. 




8. Given c 


= 72.15, A = 


: 39° 34' , 


required B = 


50° 26', a = 


= 45.958, 


6 = 55.620. 






B 


= (90° - A) = 


: 50° 26'. 


a 


= c sin A. 




log a 


= log c 4- log 


sin A, 


log c 


= 1.85824 




log sin ^L 


= 9.80412 




log a 


= 1.66236 




a 


= 45.958. 




b 


= c cos A. 




log 6 


= log c 4- log 


cos A. 


logc 


= 1.85824 




log cos A 


= 9.88699 





log 6 
6 



1.74523 
= 55.620. 



9. Given c = 1, A = 36° ; required 
B = 54°, a = 0.58779, 6 = 0.80902. 



TEACHERS EDITION". 



33 



B = (90° - A) = 54°. 


11. Given c = 93.4, 5 = 76° 25' ; 


• A ° 

sin A = - • 

c 


required A = 13° 35', a = 21.936, 

6 = 90.788. 


.-. a = csin J.. 


A = (90° -B) = 13° 35'. 


log a = log c + log sin A . 


^ = c sin ^4. 


log c = 0.00000 


log a = log c 4 log sin A. 


log sin A = 9.76922 


log c = 1.97035 


loga = 9.76922 - 10 


log sin A = 9.37081 




log a = 1.34116 


a = 0.58779. 




5 


a = 21.936. 


cos A = -• 


b — a cot A. 


.-. & = c cos ^4. 


log 6 = log a + log cot A. 


log 6 = log c 4 log cos A. 


loga = 1.34116 




' log cot 4 = 0.61687 


log c = 0.00000 


log 6= 1.95803 


log cos A = 9.90796 




log 6 = 9.90796 - 10 


& = 90.788. 


6 = 0.80902. 


12. Given a = 637, .1 = 4° 35'; 




required 5 = 85° 25', b = 7946, 




c = 7971.5. 


10. Given c = 200, B = 21°47'; 




required A = 68° 13', a = 185.72, 


J B = (90 o -^4)=85°25 / . 


h = 74.22. 


b = a cot .A. 




log b = log a 4- log cot ^4. 


^4 = (90° - B) = 68° 13'. 


log a = 2.80414 


sin ^4 = - • 


log cot A = 1.09601 


c 


log b = 3.90015 


.-. a = csin ^4. 






b = 7946. 


log a = log c 4 log sin .4. 


log c = log a 4- colog sin A. 


log c = 2.30103 


log a = 2.80414 


log sin A =9.96783 


colog sin A = 1.09740 


log a = 2.26886 


log c = 3.90154 


a = 185.72. 


c = 7971.5. 


b 
cos a — - • 


13. Given a = 48. 532, A = 36° 44'; 


c 


required B = 53° 16', B = 65.031, 


.-. 6 = c cos ^4. 


c = 81.144. 


log 6 = log c 4- log cos A . 


B = 90° -A 


log c = 2.30103 


= go - 36° 44' 


log cos A = 9.56949 


= 53° 16 / . 


log b = 1.87052 


a 


b = 74.220. 


sin A = - • 
6 



34 



PLANE TRIGONOMETRY.. 



log C 

log a ■■ 
colog sin A ■ 

log C: 
C : 

COS^i : 

.'. b : 

lOg b: 

l0gC: 

log COS A : 

lOg 6 : 

b-. 



sin^L 
= log a -f colog sin A. 
: 1.68603 
: 0.22323 
: 1.90926 

: 81.144. 
b 
c 

: C COS A. 

■ log c + log cos A. 
: 1.90926 
: 9.90386 



: 1.81312 
: 65.031. 



14. Given 
required B = 
0.000802. 

B 



sin A = - 



a = 0.0008, A =86°; 
4°, 6 = 0.0000559, c = 

= 90° - A 
= 90° - 86° 
= 4°. 

a 

c' 
a 



log c 

log a 

colog sin A 

log c 

c 

cos A 

.:b 

log 6 

log c 

log cos ^4 

log b 

b 



sin J. 
= log a + colog sin A. 
= 6.90309 - 10 
= 0.00106 
= 6.90415 - 10 
= 0.000802. 
_b 
~ c 

= c cos J.. 
= logc 4- log cos A. 
= 6.90415 - 10 
= 8.84358 - 10 
= 5.74773 - 10 
= 0.0000559. 



15. Given 6 = 50.937, B = 43° 48'; 
required A = 46° 12', a = 53.116, 
c = 73.59. 

A = (90° - B) = 46° 12'. 

tan A = - . 
6 

.-. a = 6 tan J.. 

log 6 = 1.70703 

log tan A = 0.01820 

loga= 1.72523 

a = 53.116. 

a 
sin A = - . 

c 

.-. e = 



sin ^L 

log a = 1.72523 

colog sin A = 0.14161 

log c = 1.86684 

c = 73.593. 

16. Given b = 2, £ = 3° 38' ; re- 
quired ^L = 86° 22', a = 31.496, c 
= 31.559. 

^L = (90° -B) = \ 

tan ^1 = - . 
& 

.*. a = 6 tan ^i. 

log 6 = 0.30103 

log tan A = 1.19723 

log a = 1.49826 

a = 31.496. 

a 



6° 22'. 



sin A = 



a 



sin .A 
loga = 1.49826 
colog sin A = 0.00087 
log c = 1.49913 

c= 31.559. 



TEACHERS EDITION. 



35 



17. Given 
required A = 
c = 4193.5. 
A 

sin A 
log c 
log a : 



colog sin A 

loe c = 3.62258 



a = 992, if = 76° 19' ; 
= 13° 41', b = 4074.5, 

= 90° - 76° 19' 
= 13° 41'. 

_ a 

~ c 

= log a + colog sin J.. 

= 2.99651 
= 0.62607 



sin I? : 

/. b: 

lOg b : 

lOg C: 

log sin 5 : 

lOg 6: 



: 4193.5. 

6 

c 
: c sin 5. 
: log c + log sin B. 

: 3.62258 
: 9.98750 
; 3.61008 

: 4074.5. 



18. Given a = 73, B = 68° 52' ; 
required A = 21° 8', 6 = 188.86, 
c = 202.47. 

A = (90° - £) = 21°8 / . 

sin A — -- 

c 

a 



sin J. 
log c = log a -f colog sin A. 

log A = 1.86332 

colog sin A = 0.44305 

logc = 2.30637 

c = 202.47. 

c 
.-. b = c sin Z>\ 
log 6 == log c + log sin B. 



logc = 2.30637 

log sin 5 = 9.96976 

log b = 2.27613 

b = 188.80. 



19. Given 
required A - 
c = 3.1185. 

A 



a = 2. 189, £ = 45° 25'; 
= 44°35 / , b = 2.2211, 



sin A — - 




' = 0.34657 
= 2.2211. 

20. Given b = 4, ^. = 37° 56' ; re- 
quired B = 52° 4', a =3.1176, c 
= 5.0714. 

£ = (90° - Jl) = 52° 4'. 

cos A = - • 



cos ^4 

log c = log 6 + colog cos A. 

log 6 = 0.60206 

colog cos ^. = 0.10307 

log c = 0.70513 

c = 5.0714. 



36 



PLANE TRIGONOMETRY. 



tan A = - • 
b 


b- * . 


cot^L 


.-. a = 6 tan .4. 


log 6 = log a + log cot A. 


log a = log b + log tan A. 


loga = 1.85600 


log 6 = 0.60206 


log cot A = 9.82817 


log tan ^i = 9.89177 


log 6 = 1.68417 


log a = 0.49383 


b = 48.324. 


a = 3.1176. 






23. Given c = 9.35, a = 8.49 ; re- 


21. Given c = 8590, a = 4476 ; 


quired ^. = 65° 14', B = 24° 46', 


required A = 31° 24', B = 58° 36', 


6 = 3.917. 


6 = 7332.8. 


sm A = - . 


a 
sin ^L = - . 

c 


c 
colog c = 9.02919 - 10 


log sin ^L = log a 4- colog c. 


loga = 0.92891 


log a = 3.65089 


log sin A = 9.95810 - 10 


colog c = 6.06601 - 10 


A = 65° 14'. 


log sin A = 9.71690 - 10 


.-. B = 24° 46'. 


A = 31° 24'. 


A b 
cos A = - • 


.-. B = 58° 36. 


c 


1) 


.-. 6 = c cos a. 


cot A = - . 
a 


logc = 0.97081 


6 — a cot a. 


log cos A = 9.62214 


log b — log a + log cot A. 


log b = 0.59295 


log a = 3.65089 


6 = 3.917. 


log cot A =0.21438 




log b = 3.86527 


24. Given c = 2194, b = 1312.7 ; 


b = 7332.8. 


required A = 53° 15', B = 36° 45', 




a = 1758. 


22. Given c = 86.53, a = 71.78; 


cos A—-. 


required A = 56° 3', J5 = 33° 57', 


c 


b = 48.324. 


log J? = 3.11816 


• A a 

sin A = - . 


colog c = 6.65876 -10 


c 


log cos A =9.77692-10 


log sin J. = log a + colog c. 


A = 53° 15'. 


loga = 1.85600 


.-. B = (90° - A) 


colog c = 8.06283 - 10 


= 36° 45'. 


log sin A =9.91883- 10 


a 




sm J. = - • 


A = 56° 3'. 


c 


.-. B = 33° 57'. 


,\ a = c sin -4. 



TEACHERS EDITION. 



37 



logc 
log sin A 



: 3.34124 
9.90377 



log A =3.24501 
a = 1758. 



25. Given c 

required A = 
a = 24.67. 

cos ^4 = 



= 30.69. b= 18.256; 
53° 30'. 5 = 36° 30'. 



log cos A 

log b 
colog c 
log cos A = 
A = 
..B = 

tan A — 

a = 

log a = 

log tan A — 

log b = 

log a 



=log.B -f colog c. 
= 1.26140 
. 8.51300- 10 
= 0.77440 - 10 
= 53° 30'. 
= 36° 30'. 
a 

b 

h tan ^4. 

log tan A + log b. 

0.13079 

1.26140 



log a = 1.58335 
colog sin A = 0.05012 
logc = 1.63347 
c = 43. 

27. Given a = 1.2291, b = 14.950; 
required A = 4° 42'. B = 85° 18', 
c = 15. 

a 

V 

log a = 0.08959 
colog 6 = 8.82536 - 10 
log tan A = 8.91495 
.4 = 4° 42 / . 
.-. B = 85° 18'. 



tan A 



10 



sin A = 



a 



1.39219 
24.671. 



26. Given a = 38.313, 6 = 19.- 322: 
required A = 63°, £ = 27°, c = 43. 

tan A = - ■ 



sin ^4 
log « = 0.08959 
colog sin J. = 1.08651 
logc = 1.17610 
c = 15. 

28. Given a =415.38, 6 = 62.080; 
required A = 81° 30', B = 8° 30', 
c = 420. 

a 

V 

loga= 2.61845 
colog 6= 8.20705-10 



tan ^4 



log tan A = log a -f colog /;. 


log tan A = 10.82550 - 


-10 


loga= 1.58335 


.4 = 81° 30'. 




colog 6= 8.70948 - 10 


.-. B = 8° 30'. 




log tan A = 10.29283 - 10 
A = 63°. 


sin j1 = -• 

c 




.-. B = 27°. 


a 




4 ° 

sin A = - • 
c 


sin J. 

log a = 2.61845 




Of 


colog sin A = 0.00480 
log c = 2.62325 




sin ^4 




log c = log a + colog sin .4. 


c = 420. 





38 



PLANE TRIGONOMETRY. 



29. Given a = 13.690, 6 = 16.926; 


31. Compute the unknown parts 


required A = 38° 58', B = 51° 2', 


and also the area, having given 


c = 21.769. 


a = 5, 6 = 6. 


A a 

tan A = - • 
6 


A a 

tan A — -- 
6 


log tan ^L = log a -f colog 6. 


. log tan ^4 = log a + colog -6. 


log a = 1.13640 


log a = 0.69897 


colog 6 = 8.77144 -10 


colog 6 = 9.22185- 10 


log tan A =9.90784- 10 


log tan 4 = 9.92082 - 10 


A = 38° 58'. 


A = 39° 48'. 


.-. 5 = 51° 2'. 


.-. B = 50° 12 r . 


. . a 
sin A = - . 
c 


sin .A = - • 

c 


a 


a 
* c — 


sin vl 


sin ^4 


log c = log a + colog sin J.. 


log c = log a 4- colog sin A . 


log a = 1.13640 


log a = 0.69897 


colog sin A =0.20144 


colog sin A =0.19375 


log c = 1.33784 


log c = 0.89272 


c = 21.769. 


c = 7.8112. 


30. Given c = 91.92, a = 2.19; 


2 2 


required A = 1° 22', B = 88° 88', 
6 = 91.894. 


32. Compute the unknown parts 


a 


and also the area, having given 


sin A = - • 
c 


a = 0.615, c = 70. 


log sin A = log a + colog c. 


sin A = - • 


log a = 0.34044 
colog c = 8.03659 - 10 
log sin A =8.37703 - 10 
A = 1° 22'. 


c 
log sin A = log a -f colog c. 
log a = 9.78888 -10 
colog c = 8.15490 - 10 


... B = 88° 38'. 
b 


log sin ^1 = 7.94378 -10 
A = 30' 12". 


cos ^1 = - • 

c 


.-. B = 89° 29' 48". 


.-. Z> = c cos A. 


6 = V(c 4- a) (c — a). 


log 6 = log c + log cos A. 


lotr6 _ lo g( c + a ) + lo s( c - a ) 


log c = 1.96341 


& 2 


log cos A = 9.99988 


log(c + a) = 1.84890 


log b = 1.96329 


log(c -a) = 1.84126 


6 = 91.894. 


3.69016 



teachers' edition 



39 



log 6 = 1.84508. 

b = 69.997. 

F = i ab. 
log a = 9.78888 - 10 
log b = 1.84508 
colog 2 = 9.69897 - 10 
log.F = 1.33293 

F = 21.525. 

33. Compute the unknown parts 
and also the area, having given 
b = V2, c = V3. 

V2 = 1.25991. 
V3 = 1.73205. 

, b 
cos A = - • 

c 

log cos ^4 = log 6 -f colog c. 
log 6 = 0.10034 
colog c = 9.76144 - 10 
log cos A = 9.86178 - 10 
A = 43° 20'. 
.-. £ = 46° 40'. 

smA = -. 

c 

.-. a = c sin ^1. 

log a = log c -f log sin A. 

log c = 0.23856 

log sin A = 9.83648 

log a = 0.07504 

a = 1.1886. 

F=iab. 
log a = 0.07504 
log b = 0.10034 
colog 2 = 9.69897 -10 
logF= 9.87435- 10 

F= 0.74876. 

34. Compute the unknown parts 
and also the area, having given 
a = 7, A = 18° 14', 



J5 = 71°46 / . 
sin A 



a 

' c 



. c = 



log c 

log a 

colog sin A 



sin A 
: log a + colog sin A. 
: 0.84510 
: 0.50461 




.-. 6 = 



log b ■■ 

log a: 

colog tan A ■. 

log b 

b 

F 

log a 

log b 

colog 2 

logF 



tan^l 

log a + colog tan A. 
: 0.84510 
= 0.48224 - 10 
= 1.32734 
= 21.249. 
= I aft. 
= 0.84510 
= 1.32734 
= 9.69897 - 10 
= 1.87141 
= 74.372. 



35. Compute the unknown parts 
and also the area, having given 
b = 12, A = 29° 8'. 

A = 29° 8'. 
.-. B = 60° 52' 

cos A = - • 



cos^i 

log c = log 6 + colog cos ^1. 

log b = 1.07918 

colog cos A = 0.05874 

log c = 1.13792 

c = 13.738. 



40 



PLANE TRIGONOMETRY. 



sin A = - 



.-. a 
log a 

logc 
log sin A 

log a = 0.82531 



a 

c " 

c sin J.. 

logc + log sin A. 

1.13792 
9.68739 



: 6.6882. 
\ab. 



a 
F 

log F = loga+log6+colog2. 



log a: 

log 6 = 

colog 2 = 

lOgF: 

F-. 



0.82531 
1.07918 
9.69897 - 10 
: 1.60346 

: 40.129. 



36. Compute the unknown parts 
and also the area, having given 
c = 68, A = 69° 54'. 

A = 69° 54'. 

.-. B = 20° 6'. 

. a 
sm A = ~ > 
c 

.-. a . = c sin ^4. 

log a = log c -f log sin A. 

log c = 1.83251 

log sin A = 9.97271 

log a = 1.80522 

a = 63.859. 

A b 

cos A = - . 
c 

.-. b = ccobA. 

log b = log c + log cos A. 

log c = 1.83251 

log cos A = 9.53613 

log b = 1.36864 



b: 

F-- 



-. 23.369. 
-.lab. 



loga = 1.80522 
log 6 = 1.36864 
colog 2 = 9.69897 - 10 
log F= 2.87283 
F= 746.15. 

37. Compute the unknown parts 
and also the area, having given 
c =27, B = 44° 4'. 

A = 45° 56'. 
a— c sin A. 
log a = log c -j- log sin A. 
log c = 1.43136 
log sin A = 9.85645 
log a = 1.28781 
a = 19.40. 
6 = c cos .A. 
log b = log c + log cos A. 
log c = 1.43136 
log cos A = 9.84229 
log 6 = 1.27365 
b= 18.778. 
F = lab. 
loga = 1.28781 
log 6 = 1.27365 
colog 2 = 9.69897 - 10 
log F = 2.26043 
F= 182.15. 

38. Compute the unknown parts 
and also the area, having given 
a = 47, B = 48° 49'. 

A = 41° 11'. 

b — acot^L. 

log b = log a -f log cot ^4 . 

loga= 1.67210 

log cot A = 10.05803 

log 6= 1.73013 

b = 53.719. 



TEACHERS 7 EDITION. 



41 



sin .A 
log c = log a + colog sin A. 
loga = 1.67210 
colog sin a = 0.18146 
log c= 1.85356 
c = 71.377. 
F=\ab. 
loga = 1.67210 
log b = 1.73013 
colog 2 = 9.69897 - 10 
logF = 3.10120 
F =1262 A. 

39. Compute the unknown parts 
and also the area, having given 
6 = 9, £ = 34° 44'. 

A = 55° 16'. 
a = b tan A . 
log a = log 6 + log tan A. 
log 6= 0.95424 
log tan A = 10.15908 
loga= 1.11332 
a= 12.981. 
a 



sin A 

log c = log a -f colog sin ^4. 
log a = 1.11332 
colog sin A = 0.08523 
log c = 1.19855 
c = 15.796. 
F = i ab. 
log a = 1.11332 
log b =0.95424 
colog 2 = 9.69897 - 10 
log F= 1.76653 
F= 58.416. 

40. Compute the unknown parts 
and also the area, having given 
c = 8.462. B = 86° 4'. 



A: 

a ■ 
log a: 

lOgC : 

log sin A : 
log a: 

a : 

b: 
lOg b: 
lOgC : 

log cos A -. 

lOg b: 

b-. 

F-- 

loga = 

log 6 = 

colog 2 = 

logF = 

F-- 



: 3° 56'. 

: c sin ^4. 

: logc -f log sin A. 

: 0.92747 

: 8.83630 - 10 

: 9.76377 - 10 

: 0.58046. 

:C COS ^4. 

: lOgC + lOg COS A. 

: 0.92747 
: 9.99898 
: 0.92645 
: 8.442. 

iab. 

9.76377 - 10 

0.92645 

9.69897 - 10 

0.38919 

2.4501. 



41. Find the value of F in terms 
of c and ^4. 





F = 


= -J- ab. 






sin 
cos 


^4 = 

. a - 
A = 

\ 6 = 


a 

" c 

= c sili ^4. 

6 

c 
= c cos ^4. 










= ±c 2 s'mA 


cos 


.1. 



42. Find the value of F in terms 
of a and ^4 . 

F = ±ab. 

cot A = - • 

a 

.*. b = a cot A. 
F = ±ab 
= £ a 2 cot A. 



42 



PLANE TRIGONOMETRY. 



43. Find the value of F in terms 
of b and A. 



tan A 



F = iab. 
a 
b' 

.-. a = b taxi A. 
F = iab 
= ib 2 t2Li\A. 

44. Find the value of F in terms 
of a and c. 

F = i ab. 

c 2 = a 2 4- 6 2 . 

62 - C 2 + a 2. 



a* 



.-. b= Vc 2 " 
F = iaVc 2 -a 2 . 

45. Given F = 58, a = 10 ; solve 
the triangle. 



6 = 

log 6 = 
log 2 F = 
colog a = 

log 6 = 
b = 

tan- .4 = 

log tan ^L - 

log a 
colog b = 
log tan A — 
A = 

c = 
log c = 



iab. 
2F 

a 
log 2 F 4- colog a. 
2.06446 
9.00000 - 10 
1.06446 
11.6. 
a 
7' 

, 4- colog b. 
^0000 
8.93554 - 10 
9.93554 - 10 
40° 45' 48". 
49° 14' 12". 
a 



sin A 

log a -f colog sin A . 



loga = 1.00000 

colog sin A = 0.18513 

log c- 1.18513 

c - 15.315. 



46. Given 
the triangle. 
F 



log a ■■ 

lOg 2 jP : 
COlog b : 

log a : 

a - 



tan ^t = - . 



F = 18, 6 = 5; solve 

= ^-a6. 
_2F 
b ' 
= log 2 F + colog 6. 
= 1.55630 
= 9.30103 - 10 
= 0.85733 

= 7.2. 
a 



log tan A ■. 

log a: 

colog b -- 

log tan A : 



log a 4- colog £. 
0.85733 
9.30103 - 10 



10.15836 - 

: 55° 13' 20" 

34° 46 / 40" 



10 



_ a 
sin A 
log c = log a 4- colog sin -4. 
log a = 0.85733 
colog sin A = 0.08546 
log c = 0.94279 
c = 8.7658. 

47. Given F = 12, A = 29°; solve 
the triangle. 

B = 61°. 
F = ia&=12. 
a& = 24. 
24 

a 

6* 



tan>4 



teachers' edition. 



43 



tan 29° = — 



24 
62* 



b* = - 



^4 



log 6 = i (log 
log 24 
colog tan 29° 
S 

log 6 : 

6 

tan 29° 

.*. a 

log a 

log b 

log tan 29° 

log a 

a 

sin^l 



tan 29° 
;24 + colog tan 29°). 
= 1.38021 
= 0.25625 
Q 1.63(346 
= 0.81823 
= 6.58. 
_ a 

~v 

= b tan 29°. 

= log b + log tan 29°. 

= 0.81823 

= 9.74375 

= 0.56198 

= 3.6474. 

_ a 

~ c 
a 



lOgC: 

log a ■■ 

colog sin 29° : 

logc : 



sin 29° 

log a + colog sin 29° 
: 0.56198 

0.31443 
: 0.87641 

7.5233. 



a — 



48. Given F = 100, c = 22 ; solve 
the triangle. 

F=i«&= 100. 
ab = 200. 
200 
6 

, 40000 

a 2 = 

6 2 

a 2 + 6 2 = c 2 = 484. ' 

Substitute, 

40000 



6 2 



■ + 6 2 = 484. 



40,000 + 6 4 = 484 6 2 . 



&4 _ 4 8 4 b 2 . 

64-() + (242) 2 : 

log Vl8564 : 

But 2.13434: 

.-. 62 _ 242 : 

62: 
l0£6 : 



cos A 

log cos A 
log 6 

COlCg C : 

log cos ^4 = 

A = 

/. J5 = 

sin A = 

.-. a = 

loga = 

logc = 

log sin A = 

log a: 

a 



= - 40,000. 
= 18,564. 

= { (4.26867) 
= 2.13434. 
= log 136.25. 
= 136.25. 
= 378.25. 
3 I log 378.25 
= 1.28889. 
= 19.449. 

6 
" c' 

- log 6 + colog c. 
= 1.28889 
= 8.65758 
= 9.94647 
= 27° 52'. 
= 62° 8'. 

a 

c" 

csin^L. 

logc + log sin A. 

1.34242 

9.66970 



: 1.01212 
: 10.283. 



49. Find the angles of a right 
triangle if the hypotenuse is equal 
to three times one of the legs. 

Let c = hypotenuse, 

and c = three times a, one 

of the legs, 

. . a 1 
sin A = - = - • 
c 3 

log sin A = log a + colog c. 

log a = 0.00000 

colog c = 9.52288 - 10 
log sin A = 9.52288 

^ = 19° 28' 17". 
.-. B = 70° 31' 43". 



44 



PLANE TRIGONOMETRY. 



sin^L 



50. Find the legs of a right tri- 
angle if the hypotenuse is 6, and 
one angle is twice the other. 

Let c = hypotenuse = 6, 

and B- 2 A. 

Then B = 60°, and A = 30°. 
a 
c 

.-. a = csinA. 

log a = log c + log sin .4. 

log c = 0.77815 

log sin ^. = 9.69897 

log a = 0.47712 

a = 3. 

sin B = - • 
c 

.-. 6 = c sin B. 

log 6 = log c + log sin 2?. 

log c = 0.77815 

log sin 5 = 9.93753 

log 6 = 0.71568 

6 = 5.1961. 

51. In a right triangle given c, 
and A — nB ) find a and 6. 

J5 = 90°-^. = 90°-nI?. 
£ (n + 1) = 90°. 
90° 



.'. £: 



n + 1 



cos - 



cos B = - • 
c 

90° a 



n + 1 



:. a = c cos - 



sin JB : 
90° 

L : 

n + 1 



90° 

n + 1 



.-. 6 = c sin 



90° 
n + l" 



52. In a right triangle the differ- 
ence between the hypotenuse and 
the greater leg is equal to the differ- 
ence between the two legs. Find the 
angles. 

c — a = a - b. 
2a-b = c. (1) 

a 2 + 6 2 = c 2 . (2) 

Square (1), 

4 a 2 - 4 a& + 6 2 = c 2 

a 2 + 6 2 = c 2 

3 a 2 - 4 a& = 

3 a 2 = 4 a&. 
3 a = 4 6. 
46 
3 ' 
a_4 
6~3* 

log tan A = log 4 + colog 3. 
log 4= 0.60206 
colog 3 = 9.52288 - 10 
log tan A = 10.12494 
A = 53° V 48". 
.-. B = 36° 52' 12". 

53. At a horizontal distance of 
120 feet from the foot of a steeple, 
the angle of elevation of the top 
was found to be 60° 30'. Find the 
height of the steeple. 



a = 



tan A — 



tan .4. 



a 
V 

.\ a = 6 tan A. 
log a = log b + log tan A. 
log 6= 2.07918 
log tan A = 10.24736 
loga= 2.32654 
a = 212.1. 
Height of steeple is 212.1 feet. 



TEACHERS EDITIOX. 



45 



cot A — 



54. From the top of a rock that I 
rises vertically 326 feet out of the 
water, the angle of depression of a 
boat was found to be 24°. Find the 
distance of the boat from the foot of 
the rock. 

b 
a 

.-.b — a cot A. 
log b = log a + log cot A. 
\oga= 2.51322 
log cot A = 10.35142 
log 6= 2.86464 
b = 732.22. 
Distance is 732.22 feet. 

55. How far is a monument, in 
a level plain, from the eye, if the 
height of the monument is 200 feet 
and the angle of elevation of the 
top 3° 30' ? 

cot A = - • 
a 

.-. b = a cot A. 

log b = log a + log cot A. 

log a = 2.30103 

log cot A = 1.21351 
log 6 = 3.51454 
b = 3270. 
Monument is 3270 feet from the 
eye. 

56. A distance AB is measured 
96 feet along the bank of a river 
from a point A opposite a tree C on 
the other bank. The angle ABC 
is 21° 14'. Find the breadth of the 
river. 

tan 5 = AC + AB. 
.-. AC = AB x tan B. 
log A C = log AB + log tan B. 



\ogAB= 1.98227 
log tan B = 9.58944 
log^lC = 1.57171 
^LC = 37.3. 
Breadth of river is 37.3 feet. 

57. What is the angle of eleva- 
tion of an inclined plane if it rises 
1 foot in a horizontal distance of 
40 feet ? 

a 
tan A = - • 
b 

log tan A — log a + colog b. 

log a = 0.00000 

colog b = 8.39794 - 10 

log tan A = 8.39794 

^1 = 1° 25' 56". 

58. Find the angle of elevation 
of the sun when a tower 120 feet 
high casts a horizontal shadow 70 
feet long. 

a 
b' 
120 



tan A = ■ 



tan A = 



70 



log tan A = log 120 + colog 70. 

log 120= 2.07918 
colog 70 = 8.15490-10 
log tan A = 10.23408 

A = 59° 44 x 35". 

59. How high is a tree that casts 
a horizontal shadow 80 feet in length 
when the angle of elevation of the 
sun is 50° ? 

tan A = - . 

b 

.-. a — b tan A. 
log a = log b + log tan A. 



46 



PLANE TRIGONOMETRY. 



log b = 1.90309 

log tan A = 10.07619 

loga = 1.97928 

a = 95.34. 

Tree is 95.34 feet high. 

60. A ship is sailing due north- 
east at a rate of 10 miles an hour. 
Find the rate at which she is moving 
due north, and also due east. 

Let AB be the direction of the 
vessel, and equal 1 hour's progress 
= 10 miles. 

AC = distance due east passed 
over in 1 hour. 

As the direction of the ship is 
northeast, 

A = 45°. 
b = c cos A. 
log b = log c + log cos A. 
logc = 1.00000 
log cos A = 9.84949 
log b = 0.84949 
b = 7.0712. 
AP = AC. 
Ship moves 7.0712 miles due north, 
and also due east. 

61. In front of a window 20 feet 
high is a flower-bed 6 feet wide. 
How long is a ladder that will just 
reach from the edge of the bed to 
the window ? 

tan A = - • 
b 

log tan A = log 20 -f colog 6. 

log 20= 1.30103 
colog 6 = 9.22185 - 10 
log tan A =10.52288 



A = 73° 18'. 
a 



sin A 
log c = log 20 4- colog sin A. 
log a = 1.30103 
colog sin A = 0.01871 
log c = 1.31974 
c = 20.88. 
Length of ladder is 20.88 feet. 

62. A ladder 40 feet long may be 
so placed that it will reach a window 
33 feet high on one side of the street, 
and by turning it over without mov- 
ing its foot it will reach a window 
21 feet high on the other side. Find 
the breadth of the street. 



cos B = 

log 33 = 

colog 40 = 

log cos B = 

B = 

tani? = 



log 33 : 

log tan B ■■ 
logb ■■ 

b: 
COS B' : 

lOg 21 : 

COlOg 40 : 

lOg COS B' : 

B' . 

tan B' : 

V ■ 



33 

40* 

1.51851 

8.39794 - 10 

9.91645 - 10 

34° 24' 45". 
b_ 

33" 
: 33 tan B. 
: 1.51851 
: 9.83571 
: 1.35422 

; 22.606. 

21 

40* 
; 1.32222 

: 8.39794 ■ 



10 



9.72016 
: 58° 19' 54" 
V_ 
21 ' 
21 tan B'. 



TEACHERS EDITIOX. 



47 



log 21 = 1.32222 

log tan B' = 0.20982 

log&' = 1.53204 

&' = 34.044. 

Width of street 

= 32.606 feet + 34.044 feet 
= 56.65 feet. 

63. From the top of a hill the 

angles of depression of two succes- 
sive milestones, on a straight level 
road leading to the hill, are observed 
to be 5° and 15°. Find the height 
of the hill. 




5280 



sin 5° 



a 
5280* 
5280 sin 5°. 



.-. a = 

log 5280 = 3.72263 
log sin 5° = 8.94030- 10 



log a = 2.66 
sin 10° : 



a 



loga = 

colog sin 10° = 

log c = 

cos 75° = 

.-.6 = 
log c = 



sin 10° 
2.66293 
0.76033 
3.42326 

b 

c 

c cos 75°. 

3.42326 

9.41300 - 



10 



log cos 75° 

log b = 2.83626 

6 = 685.9 
Height of hill is 685.9 feet. 



64. A fort stands on a horizontal 
plain. The angle of elevation at a 
certain point on the plain is 30°, 
and at a point 100 feet nearer the 
fort it is 45°. How high is the fort ? 

B 




Let B represent the fort, AC the 
horizontal plain, BC a ± from fort 
to plain. 

BA C = angle made by line from 
eye of observer = 30°. 

BA'C = 45° = angle of elevation 
100 feet nearer. 
From A' draw A'N _L to AB. 
In rt. A AA'N, 
Z NAA' = 30°, 
and Z NA'A = 60°. 

.-. NA' = 50 feet. 
.-. AN = V(100)2 - (50)2 
= V7500 = 50 V3 
= 86.602. 
In rt. A BNA% 
BX 
NA' 
and BN= NA' cot 16°. 

log NA'= 1.69897 
log cot 15° = 0.57195 

log 5^=2.27092 

£.¥=186.60 

AN= 86.60 
AB = 273.20 



= cot NBA' = cot lo° 



48 



PLANE TKIGOXOMETRY. 



In rt. A ABC, 
Z BAC = 30°, 
and Z ABC = 60°. 
.-. BC = iAB 

= i x 273.20 feet 
= 136.60 feet. 

65. From a certain point on the 
ground the angles of elevation of 
the belfry of a church and of the top 
of a steeple were found to be 40° 
and 51° respectively. From a point 
300 feet farther off, on a horizontal 
line, the angle of elevation of the 
top of the steeple is found to be 
33° 45'. Find the distance from 
the belfry to the top of the steeple. 

A 



AD = 



ED 




= sin 33° 45'. 



300 ft. 



Draw DE _L to AB from D. 
In A BED, 

ED 

BD 

.-. ED = 300 x sin 33° 45'. 
log 300 = 2.47712 
log sin 33° 45'= 9.74474 
log ED = 2.22186 
Z EAD = 180° - 33° 45' - (180° 
- 51°) = 17° 15'. 
In A ADE, 
ED 



AD 



= sin 17° 15'. 



log .ED 
cologsin 17° 15' 
log^iD 
In A^iDC, 

DC 

AD 

r.DC 

log AD 

log cos 51° 

log DC 

In A ADC, 

AC 

DC 

..AC-- 

log DC = 

log tan 51° : 
log^LC: 

AC: 

In A FDC, 

FC 
DC' 

.-. FC = 

log DC = 

log tan 40° = 

log FC -- 

FC-- 

Distance from 
steeple is 436.79 
= 140 feet. 



sin 17° 15' 

= 2.22186 
= 0.52791 
= 2.74977 



= cos 51°. 

= AD cos 51°. 
= 2.74977 

= 9.79887 
= 2.54864 



: tan 51°. 

: DC tan 51°. 

: 2.54864 

: 10.09163 - 10 

2.64027 
: 436.79. 



= tan 40°. 

= DC tan 40°. 
= 2.54864 
= 9.92381 
= 2.47245 

= 296.79. 

belfry to top of 
feet- 296.79 fe?t 



66. The angle of elevation of the 
top C of an inaccessible fort ob- 
served from a point A is 12°. At 
a point J5, 219 feet from A and on 
a line AB perpendicular to AC, the 
angle ABC is 61° 45'. Find the 
height of the fort. 



TEACHERS EDITION". 



49 



A B 
In rt. A CAB, = cot ABC. 

AC 



AC 



AB 



cot ABC 

log AC = \ogAB + cologcot^4£C. 
A C 




log AB = 

colog cot ABC = 
log AC = 
In rt. A ADC, 

CD _ 

AC~ 

.-. CD = 

\ogCD = \ogAC 

log AC = 

log sin CAD = 

log CD = 

CD = 



2.34044 
0.26977 
2.61021 



sin CAD. 

AC sin CAD. 
+ log sin CAD. 
2.61021 
9.31788 - 10 



1.92809 
84.74. 
Height of fort is 84.74 feet. 



Exercise X. Page 33. 



1. In an isosceles triangle, given 
a and A ; find C, c, h. 

C = 180° - 2 A 
= 2(90° -A). 

1° A 

— = cos A. 

a 
c = 2 a cos A. 

h ' A 

- = sin ^4. 
a 

h = a sin A. 

2. In an isosceles triangle, given 
a and C; find A, c, h. 

C + 2 A = 180°. 

^4 = 90° - i C. 



a 


= 


cos ^4. 


c 


= 


2 a cos ^4 


h 
a 


= 


sin A. 


h 


= 


a sin ^4. 



a 
2a = 



cos ^4. 

c 
cos .4 

c 
2 cos ^4 

= sin A. 



h = a sin A. 

4. In an isosceles triangle, given 
c and C; find ^4, a, ft. 

J. = 90° - i C. 



rt 
a = 

*. 

a 



cos ^4. 

c 
2 cos A 

sin ^4. 



3. In an isosceles triangle, given 
c and A ; find C, a, h. 

C = 180° - 2 ^4 
= 2(90° - A). 



h = asin^l. 

5. In an isosceles triangle, given 
h and A ; find C, a, c. 

C = 2(90°-^4). 

• ^ h 

smA =— ■ 



50 



PLANE TRIGONOMETRY. 









sin^l 


COS 


A 


= 


a 2a 


.* 


. c 


— 


2 a cos A. 



6. In an isosceles triangle, given 
ft and C ; find A, a, c. 

A = 90° - i C. 
ft 



sin A = 



.-. a : 



ft 



sin J. 

c 

2a 
.-. c = 2 a cos A. 



i c 

cos A = ■£- 

a 



7. In an isosceles triangle, given 
a and ft; find -4, C, c. 

sin J. = ft -i- a. 

(7 = 2(90° -4). 



cos^L 



2 < 



c 
a 2a' 

.-. c = 2 a cos J.. 



8. In an isosceles triangle, given 
c and ft ; find A, C, a. 

■ i- a h 
tan A = - — 

\c 
= 2 (90° - A). 
ft 
a' 
h 
sin J. 



sin^l = 



a 



9. In an isosceles triangle, given 
a = 14.3, c = 11 ; find A, C, ft. 

cos ^4 = ^- • 

a 

log cos ^4 = log \ c + colog a. 
log |c = 0.74036 
colog a = 8.84466 - 10 

log cos A =9.58502 - 10 



sin .A 



A = 67° 22' 50". 
. C = 2 (90° - A) 
= 45° 14' 20". 
ft 
a 

.*. ft = a sin A. 
log ft = log a -f log sin A. 

toga = 1.15684 

log sin J. = 9.96524 
log ft = 1.12058 
ft = 13.2. 

10. In an isosceles triangle, given 
a = 0.295, A = 68° 10'; find c, ft, F. 



sin A = 



ft 



.-. ft = a sin J.. 
log ft = log a -f log sin ^L. 

log a = 9.46982 - 10 

log sin A = 9.96767 - 10 

log ft = 9.43749 - 10 

ft = 0.27384. 

cos A = 

a 

.-. -j-c = a cos J.. 

log i c — log a -f log cos A. 

log a = 9.46982 - 10 

log cos A = 9.57044 - 10 

logic = 9.04026 -10 

ic = 0.109713. 
.-. c = 0.21943. 

F=ich. 
2F = eft. 
log 2 F = log c -f log ft. 

log c = 9.34130 - 10 

log ft = 9.43749 - 10 

log2F= 8.77879 -10 

2^=0.060089. 
F= 0.03004. 



TEACHERS EDITION. 



51 



11. In an isosceles triangle, given 
c = 2.352, C = 69° 49'; find a, ft, F. 
i C = 34° 54' 30". 

sin|C = ^. 



sin | C 

log a = log |c + colog sin |C 

logic = 0.07041 

colog sin | C = 0.24240 

log a = 0.31281 

a = 2.055. 

cos f C = - • 

.-. A = a cos -J- (7. 
log ft = log a + log cos | C. 
log a = 0.31281 
log cos i C = 9.91385 
log ft = 0.22666 
ft = 1.6852. 
F=|cft. 
2F=ch. 
log 2 F = log c + log ft. 
log c = 0.37144 
log ft= 0.22666 
log2F = 0.59810 
2 F = 3.9637. 
F= 1.9819. 

12. In an isosceles triangle, given 
ft = 7.4847, A = 76° 14'; find a, c, F. 
ft 



sin A = 



ft 



.-. a = 

sin A 

log a = log ft + colog sin A. 

log ft = 0.87417 

colog sin A = 0.01266 

log a = 

a = 7.706, 



tan A 



ic 



logic 

log ft 

colog tan A 

logic 

ic 

c 

F 

logF 

logic 

log ft 
logF 
F 



tan^i 
= log ft + colog tan A. 
= 0.87417 
= 9.38918 - 10 
= 0.26335 
= 1.8338. 
= 3.6676. 
= |cft. 

= log i c + log ft. 
= 0.26335 
= 0.87417 
= 1.13752 
= 13.725. 



sin A 



13. In an isosceles triangle, given 
a = 6.71, h = 6.6; find .4, C, c. 

ft 
a 

log sin A — log ft + colog a. 
log ft = 0.81954 
colog a = 9.17328 - 10 
log sin A = 9.99282 - 10 
A = 79° 36 / 30". 
.-. C = 20° 47'. 

cos^L = — 
a 

.♦. -i-c = a cos A. 
log | c = log a + log cos A. 
log a =0.82672 
log cos A = 9.25617 - 10 
log|c = 0.08289 
|c= 1.2103. 
c = 2.4206. 

14. In an isosceles triangle, given 
c = 9, ft = 20 ; find A, C, a. 



tan | O 



ft 



52 



PLANE TRIGONOMETRY. 



log tan i C ■■ 

lOg i C : 
COlOg ft : 

log tan i C = 

iC = 

G = 

2A = 

r.A = 

sin A = 



■ log ic + colog ft. 

0.65321 

8.69897 - 10 
: 9.35218 

12° 40' 49". 

25° 21' 38". 
; 180° - C. 
; 77° 19' 11". 

ft 

a' 
ft 



log a. 

logh. 
colog sin A : 
log a: 
a 



sin A 
■ logh + colog sin A. 

1.30103 
: 0.01072 
: 1.31175 
: 20.5. 



.ft = 



15. In an isosceles triangle, given 
c = 147, F = 2572.5; find^i, C, a, ft. 

F=ich. 
2F 

c 
log ft = log 2 jP + colog c. 

log2F = 3.71139 
colog c = 7.83268 - 10 
log ft = 1.54407 
ft = 35. 

* A k 

tan A— 

log tan A — log ft + colog £ c. 

logh = 1.54407 
colog i c = 8.13371 - 10 
log tan J. = 9.67778 - 10 

.4 = 25° 27' 47". 
.-. C = 2 (90° - ^1) 
= 129° 4' 26". 
ft 



sin^l 
log a = log ft + colog sin -4. 



tan J. = 



log ft = 1.54407 
colog sin J. = 0.36661 
log a = 1.91068 
a = 81.41. 

16. In an isosceles triangle, given 
ft = 16.8, F = 43.68 ; find A, C, a, c. 

F = icft. 

■ ie= r 

log $ c = log F + colog ft. 
logF =1.64028 
colog ft = 8.77469- 10 
log|c = 0.41497 
lc = 2.60. 
c = 5.2. 
h_ 

log tan ^1 = log ft -f colog | c. 

log ft = 1.22531 
colog i c = 9.58503 - 10 
log tan ^4 = 10.81034 - 10 
^L = 81° 12' 9". 
i C = 8° 47' 51". 
C = 17° 35' 42". 

cos ^4 = — 
a 

log a = log I c + colog cos A . 
logic = 0.41497 
colog cos A = 0.81547 
loga = 1.23044 
a = 17. 

17. In an isosceles triangle, find 
the value of F in terms of a and c. 

F = ich. 



ft: 



4 



4 a 2 — c 2 







= i V4 a 2 - c 2 . 


c 2 ) 



TEACHERS' EDITION. 



53 



= ic V4a 2 — c-. 

18. In an isosceles triangle, find 
the value of F in terms of a and C. 

F=±ch. 
ic = a sin £ C 
A = a cos -J- C. 
F = a sin |Cxa cos | C 
= a 2 sin 1 C cos £ C. 

19. In an isosceles triangle, find 
the value of F in terms of a and A. 

F=ich. 
ic — a cos A. 
h — a sin A. 
F — a cos A xas'mA 
= a 2 sin A cos A. 

20. In an isosceles triangle, find 
the value of F in terms of h and C. 

F = ich. 
ic = h tan | C. 
F = h (h tan \ C) 
= h 2 tan i C. 

21. A barn is 40 x 80 feet, the 
pitch of the roof is 45°; find the 
length of the rafters and the area 
of the whole roof. 

40 - 2 = 20 = \ c. 

A ic 20 

cos A = — = — . 
a a 

20 = a cos A. 

20 
.-. a = 



cos A 
log a = log 20 -f cologcos^i. 

log 20 = 1.30103 

colog cos A = 0.15051 
log a = 1.45154 



a = 28.284. 
2 x 28.284 x 80 = 4525.44. 
Length of rafters is 28.284 feet ; 
area of roof is 4525.44 square feet. 

22. In a unit circle, what is the 
length of the chord corresponding 
to the angle 45° at the centre ? 

sin£C = i- C . 

logic = log a + log sin i C. 
\oga = 0.00000 
log sin i C = 9.58284 - 10 
log£c = 9.58284 - 10 
ic = 0.38268. 
c =0.76536. 

23. If the radius of a circle is 30, 
and the length of a chord is 44, find 
the angle subtended at the centre. 

siniC = ^. 

a 

log sin i C = log i c + colog a. 
logic = 1.34242 
colog a = 8.52288 - 10 
log sin£C = 9.86530- 10 
i C = 47° 10'. 
C = 94° 20'. 

24. Find the radius of a circle if 
a chord whose length is 5 subtends 
at the centre an angle of 133°. 

sin£0 = il 

a 

log a = logic -f-cologsin-J-C 

logic = 0.39794 

colog sinjC f = 0.03760 

loga = 0.43554 

a =2.7261. 

25. What is the angle at the cen- 
tre of a circle if the corresponding 
chord is equal to f- of the radius ? 



54 



PLANE TRIGONOMETRY. 



Let a = 3, then c = 2, and \ c = 1. 

sin i = - - 

log sin i C = log 1 -f colog 3. 
log 1 = 0.00000 
colog 3 = 9.52288- 10 
log sin £ = 9.52288- 10 

i C = 19° 28' 17". 
C = 38° 56' 33". 

26. Find the area of a circular 
sector if the radius of the circle is 12, 
and the angle of the sector is 30°. 




Area O = 
Area sector == 

log area sector 

= log 30 + colog 360 
+ log it + 2 log B. 

log 30 = 1.47712 

colog 360 = 7.44370 -10 

log 7t = 0.49715 

2 logE = 2.15836 

log area = 1.57633 

Area = 37.699. 



Exercise XI. Page 35. 



1. Find the remaining parts of a 
regular polygon, given n = 10, c = 1. 

180° 

"To 



iC: 



= 18°. 



ic = 0.5. 
A = 72°. 
h — \c tan .4. 
log h = log \ c + log tan A. 
log|c = 9.69897 -10 
log tan A = 10.48822 - 10 
log ft = 0.18719 

h = 1.5388. 
logr = log \c + colog cos A. 

logic = 9.69897 -10 
colog cos A = 0.51002 
logr = 0.20899 
r = 1.618. 

log h = 0.18719 

logp = 1.00000 

log2F= 1.18719 

2F= 15.388. 
F= 7.694. 



2. Find the remaining parts of a 
regular polygon, given n = 18, r = 1. 

i C = 10°. 
.4 = 80°. 
h = r sin J.. 

logr =0.00000 
log sin A = 9.99335 - 10 
log h = 9.99335 -10 

h = 0.9848. 
ic = rcosA. 

logr = 0.00000 

log cos A = 9.23967 - 10 

logic = 9.23967 -10 

ic = 0.17365. 

p = 6.2514. 

F = ihp. 

log h = 9.99335 -10 

logp = 0.79598 

log 2^ = 0.78933 

2^=6.1564. 

^=3.0782. 

3. Find the remaining parts of a 
regular polygon, given n = 20, r = 20. 



TEACHERS' EDITION. 



55 



iC= 9°. 
A = 81°. 
ft = r sin A. 
log r = 1.30103 
log sin A = 9.99462 • 
log ft = 1.29565 
h = 19.754. 
ic = r cos -4. 
logr = 1.30103 
log cos A = 9.19433 - 
logic = 0.49536 
£c = 3.1286. 
c = 6.257. 
p = 125.14. 
F=iftp. 
log ft = 1.29565 
logp = 2.09740 
log2F= 3.39305 
2 F = 2472. 
F = 1236. 

4. Find the r 

regular polygon, 

iC = 

tan i O = ^ 

logic = 

log ft = 

log tan i C = 

logic = 

ic = 

c = 



10 



10 



cos i C = 

logr = 

log ft = 

colog cos i C = 

logr = 

r = 

F = 



emaining parts of a 
, given n = 8, ft = 1. 

22° 30'. 
" c 
h 
log ft + log tan i C. 

0.00000 
9.61722 - 10 
9.61722 - 10 
0.41421. 

0.82842. 

ft 

r 

log ft + colog cos i C. 

0.00000 

0.03438 

0.03438 

1.0824. 

ihp 

3.3137. 



5. Find the remaining parts of 
a regular polygon, given n = 11, 
F=20. 

2 F = pft. 
40 = ph. 

11 

iC = 16° 22'. 

ic 



tan J C = 



A 



tan i C : 



sin 1 C : 



Substitute values of ft and c, 
p 40 _ p* 
22 * ]p ~ 880' 
logp = i (log 880 + log tan i C). 

log 880 = 2.94448 

log tan i C = 9.46788 - 10 

2 )2.41236 

logp = 1.20618 

p = 16.076. 

c = 1.4615. 

il 

r 

log r = log i c + colog sin i C. 

logic = 9.86377 - 10 

colog sin iC = 0.55008 

logr = 0.41385 

r = 2.5933. 

in h 
cos * G = - • 

r 

log ft = log r + log cos i C. 

log r = 0.41385 

log cos i C = 9.98204 

log ft = 0.39589 

ft = 2.4882. 

6. Find the remaining parts of a 
regular polygon, given n = 7, F=l. 

14 = pft. 



56 



PLANE TRIGONOMETRY. 



P 



taniCr- 
tan i C : 



7 
; 25° 43'. 

ft ' 

p 14 _ p 2 

logp = i (log 196 + log tan £ C). 

log 196 = 2.29226 
log tan i C = 9.68271 - 10 
2 )1.97497 
logp = 0.98749 
p = 9.716. 
£c = 0.694. 

taniC = ^- 
ft 

log ft = log i c + colog tan i C. 

logic = 9.84136 -10 

colog tan i C = 0.31729 

log ft = 0.15865 

ft = 1.441. 

sin£C = ^. 
r 

log r = log i c 4- colog sin i C. 

logic = 9.84136- 10 

colog sin iC = 0.36259 

log r = 0.20395 

r = 1.5994. 

7. Find the side of a regular 
decagon inscribed in a unit circle, 
} C == 18°. 

sin£C = ^. 
r 

log c = log 2 + log sin i C. 

log 2 = 0.30103 

log sin £ C = 9.48998 

logc =9.79101 -10 

c = 0.61803. 



8. Find the side of a regular 
decagon circumscribed about a unif 
circle . 

i C = 18°. 

taniO = ^. 
ft 

log ^ c = log ft -f log tan -J C. 

log ft = 0.00000 

log tan *C = 9.51178 

logic = 9.51178 -10 

ic = 0.32492. 

c = 0.64984. 

9. If the side of an inscribed 
regular hexagon is equal to 1, find 
the side of an inscribed regular 
dodecagon. 

Let O be the centre of the circle, 
BC a, side of the hexagon, and BA 
a side of the dodecagon. Also let 
OD be J_ to BA. 
Then OB = BC = 1. 

Z BOB = 15°. 
In rt. A ODB, 

sin BOB = i AB ~ OB. 
.-. AB = 20Bx sin BOD. 
\ogAB = log 2 05 + log sin BOD. 

log 2 OB = 0.30103 
log sin 15° = 9.41300 
log^iJ5 = 9.71403 -10 
^45 = 0.51764. 

10. Given n and c, and let b 
denote the side of the inscribed 
regular polygon having 2n sides; 
find b in terms of n and c. 

Let O be the centre of the circle, 
BC the side of the polygon having 
n sides, BA the side of the polygon 
having 2 n sides. Then OA is _L to 
BC at its middle point D. 



TEACHERS EDITION. 



57 



360° _ 180° 
n 
180° 
n 
The A BOA is isosceles. 

90° 
n 
ZABC^ZOBA -ZOBC 

90° \ /_ 180° \ 90° 
n 



A BOA 

2n 

Z OBC = 90° - 



2\ 
= 90° 



\ n / \ n / n 



ic 90° 

— = COS 

b n 



-. l?c = b cos 



90° 



r.b = 



ic 



90° 



2 cos 



90° 



11. Compute the difference be- 
tween the areas of a regular octa- 
gon and a regular nonagon if the 
perimeter of each is 16. 

2 n 10 



180° 



= 22° 30'. 



log h = log £ c + log cot A. 
log£c = 0.00000 
g cot A = 10.38278 - 10 
logh = 0.38278 
logF=log/i + k)gip. 
log ft =0.38278 
logip = 0.90309 
logF = 1.28587 
F= 19.3139. 
P 



log ft' — log i c' -f log cot A'. 
logic' = 9.94885-10 
log cot A' = 10.43893 - 10 
log/i' = 0.38778 
logF^log/t' + log^p. 

logh' =0.38778 

log ip = 0.90309 

logF' = 1.29087 

W = 19.5377. 

F' -F= 19.5377 - 19.3139 

= 0.2238. 

12. Compute the difference be- 
tween the perimeters of a regular 
pentagon and a regular hexagon if 
the area of each is 12. 

F = 12, n = 5. 

5 

F=$hp. 



p 

2n 10 



tan i C ■■ 



*C 



2n' 18 



180° 
n' 



:20°. 



v_ 

£ c _ 10 _ p 2 
T ~24 ~240* 

JP 

p 2 = 240 tan i C. 

log 240 = 2.38021 

log tan i C = 9.86126 - 10 

2 )2.24147 

logp = 1.12074 

p = 13.205. 

n' = 6, i C" = 30°. 

12 z>' 2 
tan£C' = — = ^— . 
24 288 

p' 2 = 288 tan £ C". 



58 



PLANE TRIGONOMETRY. 



log 288 = 2.45939 

log tan i C" = 9.76144- 10 

2 )2.22083 

log p' = 1.11042 

p' = 12.895. 

p-p' = 0.310. 

13. Find the area of the regular 
octagon formed by cutting away the 
corners of a square whose side is 1. 

♦«-!(t) : "T- 

A = 90° - 22° 30' 
= 67° 30'. 
h 



tan A — — 
he 



ft 



tan ^4 

log i c = log h -f colog tan ^L. 
log ft = 9.69897 - 10 
colog tan A = 9.61722 - 10 
log £-c = 9.31619- 10 
p — \c x 2 n = ?ic. 
F = ipft = |cx in. 
log F = log^c + login, 
logic = 9.31619- 10 
login = 0.60206 
logF = 9.91825 -10 
F = 0.82842. 

14. Find the area of a regular 
pentagon if its diagonals are each 
equal to 12. 

180° 



ZAOD 



= 36°. 



ZAOC = 180° -36° = 141°. 
Z^lOi) = i(180 o -144 o ) 
= 18° = ZCAO. 
Z GAB = 90° - ZAOD=te°. 



ZDAC = ~ AO + 18° = 72°. 

log i c ~ log 12 + log cos 72°. 

log cos 72°= ^.48998 
log 12 = 1.07918 
logic = 0.56916 

tan DAO- — • 

ic 

log ft = log i c + log tan 54°. 




logic = 0.56916 
log tan 54° = 10.13874 -10 
log ft = 0.70790 
p = \c x 2n. 
F — iph = ic x nft. 
logF = logic+logn+logft. 
logic = 0.56916 
logw = 0.69897 
log ft = 0.70790 
logF= 1.97603 
F = 94.63. 

15. Find the area of a regular 
polygon of 11 sides inscribed in a 
circle, if the area of an inscribed 
regular pentagon is 331.8. 



I J..\« HEKfl EDITION. 



59 



Let J B be a side of a regular 
inscribed penta 7 A I) the 

j alar in* . 
(»f 1 1 fldi 

Let r be the radius of the circle 
whose centre is O. and /i and h' 
the apothems of the two ]>oh_ 
ectively. 
Given F the area ol pent 
= 881.8; find F\ the area of the 
11 -sided polygon. 

Let p and // and c and ( ■' repre- 
sent the perimeters and sides <»f the 
pentagon and the ll-aided polygon 
respectively. 

F= *j>k 
881.8 \ph. 
ph - 868.6. 



// = 


P 


6 


= 


P 

5* 




frc = 


P 
10" 




40J? = 


86°. 





tan 36° = ^=:* 



in 663 6 



P" 



" 6636 
log 7/- = log tan 86° + log 6636, 
log 6636 : 3.82191 

log tan 36° = 9.86126 - 10 

logp* = 3.68317 

log/) =1.84150. 

Since \ c = T \ of p, 

logjc = 0.84150. 

s>h\ZAOE = ^- 

r 

log r = log | r ■+- colog sin 36°. 
log|c = 0.84150 
colog sin 36° = 0.23078 
logr = 1.07237 



£AOC= = i«; : 21' 4 
82 

sin ZyiOC = ^ - r. 

logr = 1.07- 

log sin A OC = j U4gg4 - 10 

log > r' = 0.52221 

tan AOi 

h' 

+ colog tan jo/' 
log > *■' - 0.62221 
cole >C = 063221 

logA' = 1.06442 

P | y //r 

= Ic'x 11 X /<'. 
](.- ; /- 0.62221 

•l 1.041 
log A' = UVV442 
log P = 2.6U 

F = 414.07. 

16. Find the aiv.i ol ;i circle 
inscribed in an equilateral triangle 
whose perimeter Es 20. 




A Y- B 

Perimeter = 2<». 

AB= | x 20= ! 
ZOAB = lof 180° = 30°. 

r 

log^4i? = 0.52288 
log tan 30° = 0.76144 - 10 
logr = 0.28432 



tan 30° : 



60 



PLANE TRIGONOMETRY. 



Area = izr 2 . 

log it = 0.49715 

logr 2 = 0.56864 

log area = 1.06579 

Area = 11.636. 

17. Find the area of a regular 
polygon of 15 sides inscribed in a 
circle, if the area of a regular in- 
scribed polygon of 16 sides is 100. 






iC 



360° 
32 



11° 15'. 



Let 



30 

AC = h,AB = r y BC=:ic. 

F=ihp. 

100 = ihp. 

. 200 

h = 

P 

P 

oo 

* 200 

#2 = 6400 tan | C. 

log 6400 = 3.80618 
log tan i O = 9.29866 - 10 
2 )3.10484 
log i> = 1.55242 

p = 35.68. 
\c = 35.68 -32 
= 1.115. 



sin J (7 •■ 

log 1.115: 

colog sin i C ■ 
logr : 

K_ 

r 

.:h' 



1.115 
r 



: 0.04727 
: 0.70976 
: 0.75703 



= COSiC"(12 ). 

= r x cos i C. 



logr =0.75703 

log cos i C = 9.99040 - 10 . 

logft' = 0.74743 

^~ = sin i C. 

r 2 

\c' = r x sin -J- C". 

logr = 0.75703 

log sin i C = 9.31788 

log^c 7 = 0.07491 

F = -(--x 2nh'\ 
2\2 / 

log F = log i c 7 4- log n 4- log h\ 

logic 7 = 0.07491 

log 15 = 1.17609 

log 7a 7 = 0.74743 

1.99843 

F= 99.640. 

18. Find the perimeter of a 
regular dodecagon circumscribed 
about a circle the circumference 
of which is 1. 

Given circumference of inscribed 
circle = 1, n = 12 ; find p. 

2 nr = circumference. 

_ c ^ rc - 
r ~~~2~flr" 
360° 
~24~ 



iC: 



= 15°. 



tan 15° = ^ = itc. 
r 

tan 15° 



3.1416 

log tan 15° = 9.42805 
colog 3. 1416 = 9.50285 - 10 



logc = 8.93090 

log 12 = 1.07918 

logp = 0.01008 

p = 1.0235. 



10 



teachers' editiox. 



61 



19. The area of a regular poly- 
gon of 25 sides is 40 ; find the area 
of the ring comprised between the 
circumferences of the inscribed and 
circumscribed circles. 



25 

ic 

h 

jch 



1.6. 



12'. 



= tan \ C, 

= tan \ C. 

1.6 

~~ tan \ C 

log 1.6 = 0.20412 

colog tan i C = 0.89850 

log ft 2 = 1.10262 

losr/i - 0.55131. 



■ = cos -J- C. 
h 



cos i C 
\ogh = 0.55131 
colog cos £ C = 0.00344 
logr = 0.55475 
logr 2 = 1.10950. 
Ttr' 1 = area of circumscribed O. 
log ?t = 0.49715 
log r 2 = 1.10950 
logF= 1.60665 
F= 40.425. 
log 7t = 0.49715 
log h 2 = 1.10262 
log ^ 2 = 1.59977 
Area of inscribed = 39.790. 
40.425 - 39.790 = 0.635. 



Exercise XII. Page 45. 



1. Construct the functions of an 
angle in Quadrant II. What are 
their signs ? 

Sines and tangents extending up- 
wards from the horizontal diameter 
are positive ; downwards, negative. 
Cosines and cotangents extending 
from the vertical diameter towards 
the right are positive ; towards the 
left, negative. The sign of the secant 
is the same as the sign of the cosine ; 
and the sign of the cosecant is the 
same as the sign of the sine. Hence, 

sin and esc are +. 

cos and sec are — . 

tan and cot are — . 

2. Construct the functions of an 
angle in Quadrant III. What are 
their signs ? 



sin and esc are — . 
cos and sec are — . 

tan and cot are -f . 

3. Construct the functions of an 
angle in Quadrant IV. What are 
their signs ? 

sin and esc are — . 
cos and sec are -f . 
tan and cot are — . 

4. What are the signs of the 
functions of the following angles : 
340°, 239°, 145°, 400°, 700°, 1200°. 
3800° ? 

240° is in Quadrant IY. 

sin = — . tan= — . sec=+. 
cos = +. cot = — . csc = — . 



62 



PLANE TRIGONOMETRY. 



239° is in Quadrant III. 

sin = — . tan = 4- . sec = — . 

cos — — . cot = 4- . esc = — . 
145° is in Quadrant II. 

sin = + . tan = — . sec = — . 

cos = — . cot=— .. csc=+. 
400° = 360° + 40°. 
Therefore, 400° is in Quadrant I. 

sin = -f- . tan = 4- . sec = 4- . 

cos=+. *cot=-f. csc=+. 
700° = 360° + 340°. 
Therefore, 700° is in Quadrant IY. 

sin = — . 1 tan = — . sec — + . 

cos == + . cot = — . esc = — . 
1200° = 3 x 360° + 120°. 
Therefore, 1200° is in Quadrant II. 

sin = + . tan = — . sec = — . 

cos = — . cot = — . esc = + • 
3800° = 10x 360° + 200°. 
Therefore, 3800° is in Quadrant III. 

sin = — . tan = -f . sec = — . 

cos = — . cot=+. esc = — . 

5. How many angles less than 
360° have the value of the sine 
equal to -f f ? and in what quad- 
rants do they lie ? 

Since the sine is + , by Sect. XXII, 
the angles can lie in but two quad- 
rants, the first and second. 

In the first quadrant, by Sect. 
XXII, the sine increases from to 
1, and in the second, decreases from 
1 toO. 

Therefore, of angles less than 360° 
there may be two with sine equal 
to + 4 ; one in the first quadrant, 
and one in the second quadrant. 



6. How many values less than 
720° can the angle x have if cosx 
= 4- | , and in what quadrants do 
they lie ? 

720° is twice 360°; hence, the 
moving radius will make exactly 2 
complete revolutions. 

The cosine has the + sign in the 
first and fourth quadrants. 

Hence, x can have four values I 
two in Quadrant I and two in 
Quadrant IV. 

7. If we take into account only 
angles less than 180°, how many 
values can x have if sin x == f ? if 
cos x = \ ? if cos x = — f ? if tan 
x. = I ? if cot x = - 7 ? 

(i) Sign being 4-, the angle can 
be in Quadrant I or II. 

Therefore, there can be two values 
of x less than 180°. 

(ii) Sign being -j-, the angle can 
be in Quadrant I or IV. 

Therefore, there can be one value 
of x less than 180°. 

(iii) Sign being — , the angle can 
be in Quadrant II or III. 

Therefore, there can be one value 
of x less than 180°. 

(iv) Sign being -f, the angle can 
be in Quadrant I or III. 

Therefore, there can be one value 
of x less than 180°. 

(v) Sign being — , the angle can 
be in Quadrant II or IV. 

Therefore, there can be one value 
or x less than 180°. 

8. Within what limits must the 
angle x lie if cos x = — f ? if cot x 
= 4 ? if sec x = 80 ? if esc x = - 3 ? 
(If x<360.) 



TEACHERS EDITION. 



63 



If cos x = — f , x must lie in the 
second or third quadrant, or be- 
tween 90° and 270°. 

If cot x — 4, x is between 0° and 
90° or 180° and 270°. 

If sec x — 80, x is between 0° and 
90° or 270° and 360°. 

If esc x — — 3, x is between 180° 
and 360°. 

9. In what quadrant does an 
angle lie if sine and cosine are both 
negative ? if cosine and tangent are 
both negative ? if cotangent is posi- 
tive and sine negative ? 

(i) Sine is negative in Quadrants 
II and III; cosine is negative in 
Quadrants III and IV. 

Therefore, angles having both 
sine and cosine negative are in 
Quadrant III. 

(ii) Cosine is negative in Quad- 
rants II and III ; tangent is nega- 
tive in Quadrants II and IV. 

Therefore, angles having both 
cosine and tangent negative are in 
Quadrant II. 

(iii) Cotangent is positive in Quad- 
rants I and III ; sine is negative in 
Quadrants III and IV. 

Therefore, angles having cotan- 
gent positive and sine negative are 
in Quadrant III. 

10. Between 0° and 3600° how 
many angles are there whose sines 
have the absolute value § ? Of these 
sines how many are positive and 
how many negative ? 

Between 0° and 3600° there are 
10 revolutions, and in each there 
are 4 angles whose sines have the 
absolute value f . Therefore, there 



are 40 angles. The sine is positive 
in Quadrants I and II, and negative 
in Quadrants III and IV. Hence, 
there are 20 angles with the sine posi- 
tive, and 20 with the sine negative. 

11. In finding cos x b y means o f 
the equation cos x — ± vl — sin 2 x, 
when must we choose the positive 
sign and when the negative sign? 

Since cosines of angles in Quad- 
rants I or IV are positive, we use 
the sign + when angle <t lies within 
these limits. 

Also, since cosines of angles in 
Quadrants II and III are negative, 
we use the sign — , when x lies in 
either of these. 

12. Given cos x — — VX ; find the 
other functions when x is an angle 
in Quadrant II. 

sin 2 x + cos 2 x = 1. 
sin x = Vl — cos 2 x 



: Vi -(-Vi)'2 = Vi 

:W2. 



tanx 



cot x 



1 

|V2 
1 
' cos x _ i ^2 
sin x "vi 



sin x 

1 



cosx 

1 J_ 

tanx — 1 



■*vs" 






1. 



13. Given tan x = V3 ; find the 
other functions when x is an angle 
in Quadrant III. 

tan x = V§. 

cot x = -7= = i V3. 

VI 



64 



PLANE TRIGONOMETRY. 



sin x 

tan x = 

cosx 

tan x x cos x = sin x. 
V3 cos x = sin x. 
3cos 2 x — sin 2 x = 
cos 2 x + sin 2 x = 1 
4 cos 2 x = 1 

cos 2 x = i. 
cos X = ± -J-. 
The angle being in Quadrant III, 
the cosine is negative. 
.*. cos x = — i. 

sin x = Vl -(--J) 2 
= V} = ± J V3. 
Sine is negative. 

-W3. 
1 



.-. sin x • 

sec x = = — 2. 



esc x == 



iV5 



= -fV3. 



14. Given sec x = + 7, and tan x 
negative ; find the other functions 
of x. 

x must be in Quadrant IV. 

Hence, sine, cosecant, tangent, 
and cotangent are negative, and 
cosine positive. 

1 1 



sin x 



=±v* 



: - f VS. 



49 



:± 



V 



CSCX : 



sin x __ 4 V^ 



tan x = 



' 12 
sinx _ 
cosx 
= - 4 V§. 



>VI 



cot X = 



1 



1 



tan x 4 V3 

— _ _i_ vs 

— 12 v o, 

15. Given cot x = — 3 ; find all 
the possible values of the other 
functions. 

By [3] tan x = — -J-, and may be 
in Quadrant II or IV. 

By PI, 

sin 2 x = 1 — cos 2 x. 



Vl — COS 2 X 



sin x = Vl 
By [2], 

_ 1 _ 

3 cosx 

1 _ 1 — COS 2 X 

9 cos 2 x 
cos 2 x = 9 — 9 cos 2 x. 



cos^x = 



10 
3 



A v io, 



V10 

and is — in Quadrant II, -f in IV. 

By [i], 



sin x 



= V^I = \fe 



= AvTo, 



ao 

in IV. 



sec x = 



= 1 VlO. 



and is + in Quadrant II, 
VlO 
3~ 

esc x = VlO, 
and is + in Quadrant II, — in IV. 



16. What functions of an angle 
of a triangle may be negative ? In 
what case are they negative ? 

When an angle of a triangle is 
acute, its functions are all positive. 
When an angle is obtuse, its func- 
tions are those of an angle in Quad- 
rant II. 



TEACHERS EDITION. 



65 



Hence, sine and cosecant are 
always positive, and cosine, tan- 
gent, cotangent, and secant may be 
negative. 

17. Why may cot 380° be con- 
sidered equal either to + co or to 

-CO? 

The nearer an acute angle is to 
0°, the greater the positive value of 
its cotangent ; and the nearer an 
angle is to 360°, the greater the 
negative value of its cotangent. 
When the angle is 0° or 360°, the co- 
tangent is parallel to the horizontal 
diameter and cannot meet it. But 
the cotangent of 360° may be re- 
garded as extending either in the 
positive or in the negative direction, 
and hence either -f x or -x. 

18. Obtain by means of Formu- 
las [l]-[3] the other functions of 
the angles, given : 

(i) tan 90° = co. 

(ii) cos 180° = - 1. 

(iii) cot 270° = 0. 

(iv) esc 360° = - cc. 

(i) 
1 

o' 



tan 90° = oo : 



cot 90° = - = 0. 

CO 

sin 90° 1 



cos 90° 

cos 90° = sin 90° = 0, 
cos 2 90° + sin 2 90° = 1. 
sin 2 90° = 1. 
* sin 90° = 1. 

1 1 



sec 90° = - 



cos 90° 

esc 90° = — — = 1 = 1. 

sin 90° 1 





(ii) 




cos 180° 


= - l. 




sin 2 180° 


-f cos-' 180° = 


= 1. 


sin 2 180° 


+ 1 = 1. 




sin 180° 


= 0. 




tan 180° 


sin 180° 
cos 180° 
= -0. 





cot 180° 


cos 180° 
sin 180° 

= — X. 


- i 


sec 180° 


1 


i 


cos 180° 


^T 




= - 1. 




esc 180° 


1 


i 


sin 180° 


o~ 




(iii) 




cot 270° 


= 0. 




tan 270° 


1 

= - = oo. 





cos 270° 

sin 270° 


= 0. 




cos 270° 


= sin 270° 


= 0. 


sin 2 270° 


+ cos 2 270° = 


: 1. 


sin 2 270° 


+ = 1. 




sin 2 270° 


= 1. 




sin 270° 


= -1. 





sec 270° 
esc 270° = 



■ = - = CO. 



cos 270° 
1 U 



sin 270° - 1 
= -1. 

(iv) 



sin 360° = 



0. 



esc 360° = - co. 
1 

- CO 

sin 2 360° + cos 2 360°= 1. 
cos 2 360° = 1. 



66 



PLANE TRIGONOMETRY. 



cos 360°= 1. 

tan 360° = — = - 0. 
1 



cot 360° = 
sec 360° = 




>s 360° 1 



19. Find the values of sin 450°, 
tan 540°, cos 630°, cot 720°, sin 810°, 
esc 900°. 

sin 450° = sin (360° + 90°) 

= sin 90° 

= 1. 
tan 540° = tan (360° + 180°) 

= tan 180° 

= 0. 
cos 630° = cos (360° + 270°) 

= cos 270° 

= 0. 
cot 720° = cot (360° + 360°) 

= cot 360° 

= GO. 

sin 810° = sin (2 x 360° + 90°) 

= sin 90° 

= 1. 
esc 900° = esc (2 x 360° + 180°) 

= esc 180° 



20. Compute the value of 

a sin 0° + b cos 90° - c tan 180°. 

sin 0° = 0. 
cos 90° = 0. 
tanl80°=0. 
Substituting, 

ax0 + 5x0-cx0 = 0. 

21. Compute the value of 

a cos 90° - b tan 180° + c cot 90°. 

cos 90° = 0. 

tan 180° =0. 

cot 90° =0. 
Substituting, 
ax0-6x0 + cx0 = 0. 

22. Compute the value of 
a sin 90° - b cos 360° 

+ (a - b) cos 180°. 
sin 90° =1. 
cos 360° = 1. 
cos 180° = - 1. 
Substituting, 
axl-&xl + (a-&)x(-l) = 

23. Compute the value of 

(a 2 - & 2 ) cos 360° - 4 ab sin 270°. 
cos 360°= 1. 
sin 270° = - 1. 
Substituting, 
(a 2 -& 2 ) x 1 -4a& x (- 1) 
= a 2 - b 2 + 4 ab. 



Exercise XIII. Page 51. 



1. Express sin 250° in terms of 
the functions of an acute angle less 
than 45°. 

sin 250° = sin (270° - 20°) 
s= - cos 20°. 



2. Express sin 172° in terms of 
the functions of an angle less than 
45°. 

sin 172° = sin (180° -8°) 
= sin 8°. 



TEACHERS' EDITION. 



67 



3. Express cos 100° in terms of 
the functions of an angle less than 

45°. 

cos 100° = cos (90° + 10°) 
= - sin 10°. 

4. Express tan 125° in terms of 
the functions of an angle less than 
45°. 

tan 125° = tan (90° + 35°) 
= - cot 35°. 

5. Express cot 91° in terms of 
the functions of an angle less than 
45°. 

cot 91° = cot (90° + 1°) 
= - tan 1°. 

6. Express sec 110° in terms of 



11. Express cot 264° in terms of 
the functions of an angle less than 
45°. 

cot 264° = cot (270° - 6°) 
= tan 6°. 

12. Express sec 244° in terms of 
the functions of an angle less than 
45°. 

sec 244° = sec (270° - 26°) 
= - esc 26°. 

13. Express esc 271° in terms of 
the functions of an angle less than 
45°. 

esc 271° = esc (270°+ 1°) 
= — sec 1°, 

14. Express sin 163° 49' in terms 



the functions of an angle less than , of the functions of an angle less than 



45°. 



sec 110° = sec (90° +20°) 

= - esc 20°. 



45°. 



sin 163° 49' = sin (180° - 16° ll 7 ) 
= sin 16° ll 7 . 



7. Express esc 157° in terms of M. Express cos 195° 33' in terms 
the functions of an angle less than ot the functions of an angle less than 



45°. 



45°. 



esc 157° = esc (180° - 23°) 

= esc 23°. 



cos 195° 33' = cos (180° + 15° 33') 
= - cos 15° 33'. 



8. Express sin 204° in terms of 16. Express tan 269° 15' in terms 
the functions of an angle less than of the functions of an angle less than 



45°. 



sin 204° = sin (180° + 24°) 

= - sin 24°. 



9. Express cos 359° in terms of 
the functions of an angle less than 

45°. 

cos 359° = cos (360° - 1°) 
= cos 1°. 

10. Express tan 300° in terms of 18 - Express sec 299° 45' in terms 
the functions of an angle less than of the functions of an angle less than 
45°. 45°. 



45°. 

tan 269° 15' = tan (270° - 45') 
= cot 45'. 

17. Express cot 139° 17' in terms 
of the functions of an angle less than 
45°. 

cot 139° 17' = cot (180° - 40° 43') 
= - cot 40° 43'. 



tan 300° = tan (270° + 30°) 
= - cot 30°. 



sec 299° 45' = sec (270° + 29° 45') 
= esc 29° 45'. 



68 PLANE TRIGONOMETRY. 

19. Express esc 92° 25' in terms of the functions of an angle less than 

45°. 

esc 92° 25" = esc (90° + 2° 25') = sec 2° 25'. 

20. Express all the functions of — 75° in terms of the functions of a 
positive angle less than 45°. 

sin (- 75°) = sin (270° + 15°) = - cos 15°. 
cos (- 75°) = cos (270° + 15°) = sin 15°. 
tan (- 75°) = tan (270° + 15°) = - cot 15°. 
cot (- 75°) = cot (270° + 15°) = - tan 15°. 

21. Express all the functions of — 127° in terms of the functions of a 
positive angle less than 45°. 

sin (- 127°) = sin (270° - 37°) = - cos 37°. 
cos (- 127°) = cos (270° - 37°) = - sin 37°. 
tan (- 127°) = tan (270° - 37°) = cot 37°. 
cot (- 127°) = cot (270° - 37°) = tan 37°. 

22. Express all the functions of — 200° in terms of the functions of a 
positive angle less than 45°. 

sin (- 200°) = sin (180° - 20°) = sin 20°. 
cos (- 200°) = cos (180° - 20°) = - cos 20°. 
tan (- 200°) = tan (180° - 20°) = - tan 20°. 
cot (- 200°) = cot (180° - 20°) = - cot 20°. 

23. Express all the functions of — 345° in terms of the functions of a 
positive angle less than 45°. 

sin (-345°) = sin 15°. 
cos (-345°) = cos 15°. 
tan (-345°) = tan 15°. 
cot (-345°) = cot 15°. 

24. Express all the functions of — 52° 37' in terms of the functions of 
a positive angle less than 45°. 

sin (- 52° 370 = sin (270° + 37° 230 = - cos 37° 23'. 
cos (- 52° 370 = cos ( 2 ™° + 37° 230 = sin 37° 23'. 
tan (- 52° 370 = tan (270° + 37° 230 = - cot 37° 23'. 
cot (- 52° 370 = cot (270° + 37° 23') = - tan 37° 23'. 

25. Express all the functions of — 196° 54' in terms of the functions of 
a positive angle less than 45°. 



teachers' edition. 69 

sin ( - 196° 540 = sin (180° - 16° 540 = sin 16° 54'. 
cos (- 196° 540 = cos (180° - 16° 54') = - cos 16° 54 / . 
tan (- 196° 540 = tan (180° - 16° 540 = - tan 16° 54'. 
cot (- 196° 540 = cot (180° - 16° 540 = - cot 16° 54'. 

26. Find the functions of 120°. 

sin 120° = sin (90° + 30°) = cos 30° = £ \ 7 3. 
cos 120° = cos (90° + 30°) = - sin 30° = - \. 
tan 120° = tan (90° + 30°) = - cot 30° = - Vs. 
cot 120° = cot (90° + 30°) = - tan 30° = - I \ 3. 

27. Find the functions of 135°. 

sin 135° = sin (90° + 45°) = cos 45° = \ Vk 
cos 135° = cos (90° + 45°) = - sin 45° = - i V2. 
tan 135° = tan (90° + 45°) = - cot 45° = - 1. 
cot 135° = cot (90° + 45°) = - tan 45° = - 1. 

28. Find the functions of 150°. 

sin 150° = sin (180° - 30°) = sin 30° = f 
cos 150° = cos (180° - 30°) = - cos 30° = - i Vs. 
tan 150° = tan (180° - 30°) = - tan 30° = - 4- V3. 
cot 150° = cot (180° - 30°) = - cot 30° = -Vs. 

29. Find the functions of 210°. 

sin 210° = sin (180° + 30°) = - sin 30° = - £. 
cos 210° = cos (180° + 30°) = - cos 30° = - £ V3. 
tan 210° = tan (180° + 30°) = tan 30° = I Vs. 
cot 210° = cot (180° + 30°) = cot 30° = V3. 

30. Find the functions of 225°. 

sin 225° = sin (180° + 45°) = - sin 45° = - £ V5. 
cos 225° = cos (180° -f 45°) = - cos 45° = - $ V2. 
tan 225° = tan (180° + 45°) = tan 45° = 1. 
cot 225° = cot (180° -f 45°) = cot 45° = 1. 

31. Find the functions of 240°. 

sin 240° = sin (270° - 30°) = - cos 30° = - i V3. 
cos 240° = cos (270° - 30°) = - sin 30° = - £. 
tan 240° = tan (270° - 30°) = cot 30° = Vs. 
cot 240° = cot (270° - 30°) = tan 30° = i V& 



70 PLANE TKIGONOMETRY. 

32. Find the functions of 300°. 

sin 300° = sin (270° + 30°) = - cos 30° = - i Vs. 
cos 300° = cos (270° + 30°) = sin 30° = £. 
tan 300° = tan (270° + 30°) = - cot 30° = - Vs. 
cot 300° = cot (270° + 30°) = - tan 30° = - i Vs. 

33. Find the functions of - 30°. 

sin (- 30°) = - sin 30° = - j-. 
cos (- 30°) = cos 30° = i V3. 
tan ( - 30°) = - tan 30° = - J V3. 
cot ( - 30°) = - cot 30° = - V3. 

34. Find the functions of - 225°. 

- 225° = 90° + 45°. 
sin (- 225°) = sin (90° + 45°) = cos 45° = i V2 . 
cos (- 225°) = cos (90° + 45°) = - sin 45° = - i V2. 
tan(- 225°) = tan (90° + 45°) = - cot 45° = - 1. 
cot (- 225°) = cot (90° + 45°) = - tan 45° =* - 1. 

35. Given sin x = — Vj, and cos x negative ; find the other functions 
of sc, and the value of x. 

Since sin 45° = — Vj- and the sign of the cosine is negative, the angle 
must be in Quadrant III, and must be, therefore, 
180° + 45° = 225°. 
cos (180° + 45°) = - cos 45° = - i V2. 
tan (180° + 45°) = tan 45° = 1. 
cot (180° + 45°) = cot 45° = 1. 

36. Given cot x =? — v3, and x in Quadrant II; find the other func 
tions of x, and the value of x. 

Since cot 30° = V3 and the sign is negative, the angle is 180° — 30" 

= 150°. 

sin 150° = sin (180° - 30°) = sin 30° = |. 

cos 150° = cos (180° - 30°) = - cos 30° = - i V3. 

tan 150° = tan (180° - 30°) = - tan 30° = - I V3. 

37. Find the functions of 3540°. 

3540° = 9x 360° + 300°. 
sin 300° = sin (270° -f 30°) = - cos 30° = - -J- VS. 
cos 300° = cos (270° + 30°) = sin 30° = J. 
tan 300° = tan (270° + 30°) = - cot 30° = - Vs. 
cot 300° = cot (270° + 30°) = - tan 30° = - i V§. 



TEACHERS EDITION. < 1 

38. What angles less than 360° have a sine equal to — \ ? a tangent 
equal to - Vs ? 

(i) Since sin 30° = -J- and the sign is negative, the angle must be in 
Quadrant III or IV, and must be, therefore, 180° + 30° = 210°, or 360° 

- 30° = 330°. 

(ii) Since tan 60° = V3 and the sign is negative, the angle must be in 
Quadrant II or IV, and must be, therefore, 180° - 60° = 120°, or 360° 

- 60° = 300°. 

39. Which of the angles mentioned in Examples 27-34 have a cosine 
equal to — £ V2 ? a cotangent equal to — V3 ? 

(i) Since cos 45° = -J- V2 and the sign is negative, the angle must be in 
Quadrant II or III, and must be, therefore, 180° - 45° = 135°, or 180° + 45° 
= 225°. Also, the functions of — 225° are the same as the functions of 
360° - 225° = 135°. Hence, the angles are 135°, 225°, or - 225°. 

(ii) Since cot 30° = V3 and the sign is negative, the angle must be in 
Quadrant II or IV, and must be, therefore, 180° - 30° = 150°, or 360° 

- 30° = 330°, or - 30°. Hence, the angles are 150° or - 30°. 

40. What values of x between 0° and 720° will satisfy the equation 
sin x = + i ? 

Since sin 30° = i and the sign is positive, the angle must be in Quad- 
rant I or II, and must be, therefore, 30° or 180° - 30° = 150°, the first 
revolution. In the second revolution these angles must be increased by 
360°. Hence, the angles are 30°, 150°, 390°, and 510°. 

41. Find the other angle between 0° and 360° for which the corre- 
sponding function (sign included) has the same value as sin 12°, cos 2(3°, 
tan 45°, cot 72°; sin 191°, cos 120°, tan 244°, cot 357°. 

In order that the sign shall be the same, 

sin 12° must be in Quadrant II = sin (180° - 12°) = sin 168°. 
cos 26° must be in Quadrant IV = cos (360° - 26°) = cos 334°. 
tan 45° must be in Quadrant III = tan (180° + 45°) = tan 225°. 
cot 72° must be in Quadrant III = cot (180° + 72°) = cot 252°. 
sin 191° must be in Quadrant IV = sin (360° - 11°) = sin 349°. 
cos 120° must be in Quadrant III = cos (180° -f 60°) = cos 240°. 
tan 244° must be in Quadrant I = tan (244° - 180°) = tan 64°. 
cot 357° must be in Quadrant II = cot (357° - 180°) = cot 177°. 

42. Given tan 238° = 1.6 ; find sin 122°. 

tan 238° = tan (180° + 58°) = tan 58°. 
sin 122° = sin (180° - 58°) = sin 58°. 



72 PLANE TRIGONOMETRY. 

But 



tan 238° 


= 


1.6. 




tan 58° 


= 


1.6. 




tan 58° 


= 


sin 
cos 


58° 

58° 


1.6 







sin 58° 



Vl-sin 2 58° 
2.56 - 2.56 sin 2 58° = sin 2 58°. 
3.56 sin 2 58° = 2.56. 

jo ^ft 
sin58° = A -— = 0.8480. 
\3.56 

43. Given cos 333° = 0.89 ; find tan 117°. 

cos 333° = 0.89 = cos (270° + 63°) = sin 63°. 
tan 117° = tan (180° - 63°) = - tan 63°. 
sin 2 63 + cos 2 63° = 1. 
(0.89) 2 + cos 2 63° = 1. 

cos 2 63° = 0.2079. 

cos 63° =0.4559. 

_ tan63 o = _?^! = _J^_ = _ 1 . 9 5 2 2. 

cos 63° 0.4559 

44. Simplify the expression 45. Simplify the expression 

a cos (90° - x) + b cos (90° + x). m cos (90° - x) sin (90° - x). 

a cos (90° - x) + b cos (90° + x) m cos (90° - x) sin (90° - x) 

~ a sin x + b ( — sin x) = m sin x cos x. 

= (a — b) sin x. 

46. Simplify the expression 

(a - b) tan (90° - X ) + (a + b) cot (90° + x). 
tan (90° — x) = cot x. 
cot (90° + x) = - tan x. 
Hence, the expression = (a — b) cot x — (a + b) tan x. 

47. Simplify the expression 

a 2 + b 2 - 2 aft cos (180° - x). 
a 2 4- 6 2 - 2 ab cos (180° - x) = a 2 + & 2 - 2 a& (- cos x) 

= a 2 + 6 2 4- 2 a& cos x. 

48. Simplify the expression 

sin (90° + x) sin (180° + x) + cos (90° + x) cos (180° - x). 
sin (90° + x) sin (180° + x) + cos (90° + x) cos (180° - x) 

= (cos x) (— sin x) -f (— sin x) (— cos x) 

= — sin x cos x + sin x cos x = 0. 



TEACHEKS* EDITION. 73 

49. Simplify the expression 

cos (180° + x) cos (270° - y) - sin (180° + x) sin (270° - y). 

cos (180° + x) = — cos x. 
cos (270° — y) = — sin y. 
sin (180° + x) = — sin x. 
sin (270° — y) — — cos y. 
Hence, the expression = cos x sin y — sin x cos ?/. 

50. Simplify the expression 

tan x + tan (- y) - tan (180° - y). 

tan (— y) = — tan y. 
-tan (180° -y) = tan?/. 
Hence, the expression = tan x. 

51. For what values of x is the expression sin x + cos x positive, and 
for what values negative ? 

If x is in Quadrant I, sin x + cos x must be positive, since both the sine 
and cosine are positive. In Quadrant II the sine is positive and cosine 
negative; hence, so long as the sine is greater than, or equal to, the 
cosine, the expression sin x -f cos x is positive ; but after passing the 
middle of Quadrant II, viz., 135°, the cosine of x is greater than sine, 
and the expression is negative. In Quadrant III both sine and cosine 
are negative, and hence their sum must be negative. In Quadrant IV 
the sine is negative and cosine positive. The sine and cosine are equal 
at 315°, after which the cosine is greater than sine. Hence, the expression 
sin x + cos x is negative from 135° to 315°, and positive between 0° 
and 135°, and 315° and 360°. 

52. Answer the questions of Example 51 for sin x — cos x. 

As x increases from 0° to 45°, the sine increases in value, and cosine 
decreases, until at 45° sine = cosine. Hence, up to this point sin x — cosx 
is negative. For the remainder of Quadrant I sine is greater than cosine, 
and consequently the expression sin x — cos x is positive. In Quadrant II 
sine is positive and cosine negative, so the expression sin x — cos x is 
uniformly positive. In Quadrant III sine is negative and cosine negative ; 
hence, so long as sine is less than cosine, the expression is positive, viz., 
to 225° ; after that point, sine is greater than cosine, and sin x — cos x is 
negative. In Quadrant IV sine is negative and cosine positive; there- 
fore, sin x — cos x is uniformly negative. The expression is, then, 
negative between 0° and 45°, and 225° and 360°; positive between 45° 
and 225°. 



74 PLANE TRIGONOMETRY. 

53. Find the functions of x — 90° in functions of x. 

x - 90° = 360° - (90° - x) = 270° + x. 
sin (x - 90°) = sin (270° + %) = - cos x. 
cos (x - 90°) = cos (270° + x) = sin x. 
tan (x - 90°) = tan (270° + x) = - cot x. 
cot (x - 90°) = cot (270° + x) = - tan x. 

54. Find the functions of x — 180° in functions of x. 

x _ 180° = 360° - (180° - x) = 180° + x. 
sin (x - 180°) = sin (180° + x) = - sin x. 
cos (x - 180°) = cos (180° + x) = - cos x. 
tan (x - 180°) = tan (180° + x) = tan x. 
cot (x - 180°) = cot (180° + x) = xsot x. 

Exercise XIV. Page 60. 

1. Find the value of sin (x + y) and cos (x + y) when sin x = f , cos x 

= |, sin?/=: T 5 3, cos ?/ = if. 

sin (x + 2/) = sni £ cos V + cos x sin y 

56 
65* 
cos (x -f- y) = cos x cos ?/ — sin x sin ?/ 

_48_15_33 
"65 65~65* 



_/3 12\ /4 j>\_36 20 _, 

\5 X T3/\5 X T3/65 + 65~< 

= cos x cos y — sin x sin y 

\5 13/ \5 13/ 



2. Find sin (90° — y) and cos (90° — y) by making x = 90° in Formulas 
[8] and [9]. 

sin (90° — y) = sin 90° cos y — cos 90° sin y 

= (1 x cos y) — (0 x sin y) = cos y. 

cos (90° — y) — cos 90° cos y + sin 90° sin y 

= (0x cos y) + (1 x sin y) = sin y. 

3. Find, by Formulas [4]— [11], the first four functions of 90° -f y. 

sin (90° -f y) = sin 90° cos y + cos 90° sin y 

= (1 x cos 2/) + (0 x sin ?/) = cos y. 

cos (90° + y) — cos 90° cos y — sin 90° sin y 

= (0 x cos ?/) — (1 x sin y) = — sin 2/. 

tan (90° + 2/) = - -~ = - cot y. 
sin 2/ 

cot (90° + y) =■ - S1I1 - 2/ = - tan ?/. 
v cos?/ 



TEACHERS' EDITION. < O 

4. Find, by Formulas [4]-[ll], the first four functions of 180° — y. 

sin (180° - y) = sin 180° cos y - cos 180° sin y 

= (0 x cos y) — ( — 1 x sin y) = sin y. 

cos (180° - y) = cos 180° cos y + sin 180° sin y 

= (— 1 x cos y) 4- (0 x sin y) =— cos y. 

sin 2/ 
tan (180° - y) = = - tan y. 

x cosy 

cot (180° -y)=- ^^ = - cot y. 
sin y 

5. Find, by Formulas [4]— [11], the first four functions of 180° -f y. 

sin (180° + y) = sin 180° cos y + cos 180° sin y 

= (0x cos y) + ( — 1 x sin y) = — sin y. 

cos (180° -f y) = cos 180° cos y - sin 180° sin y 

= (— 1 x cos y) — (0 x sin y) = — cos y. 

tan (180° + y) = ~ Sm V = tan y. 
v ' — cos y 

cot (180° + y) = ~ cosy = cot y. 

— sin ?/ 

6. Find, by Formulas [4]-[ll], the first four functions of 270° — y. 

sin (270° — y) = sin 270° cos y — cos 270° sin y 

= (-lX cos y) — (0 x sin y) = — cos y. 

cos (270° — ?/) = cos 270° cos y + sin 270° sin y 

= (0 x cos y) + (— 1 x sin ?/) = — sin y. 

tan (270° - y) = ~ C0S V = cot y. 

— sin 2/ 

— sin v 

cot (270° - y) = ~ = tan y. 

v — cos y 

7. Find, by Formulas [4]-[ll], the first four functions of 270° -f y. 

sin (270° -f y) - sin 270° cos y + cos 270° sin y 

= (-lx cos y) + (0 x sin y) = — cos y. 

cos (270° + y) = cos 270° cos y — sin 270° sin y 

= (0 x cos y) — ( — 1 x sin y) = sin y. 

tan (270° + y) = ~ cos y = - cot y. 
sin ?/ 

cot (270° + y) = smy = - tan y. 

7 — cosy 



76 PLANE TRIGONOMETRY. 

8. Find, by Formulas [4]-[ll], the first four functions of 360° - y. 

sin (360° - y) = sin 360° cos y - cos 360° sin y 

= (0x cos y) — (1 x sin y) — — sin y. 

cos (360° -y) = cos 360° cos y + sin 360° sin y 

= (1 x cos y) + (0 x sin y) = cos y. 

tan (360° - y) = ~ Sm V = - tan y. 
V - cos?/ 

cot (360° - y) = °° Sy = - cot y. 
— sin y 

9. Find, by Formulas [4]-[ll], the first four functions of 360° -f y. 

sin (360° + y) = sin 360° cos y + cos 360° sin y 

= (0 x cos ?/) -f (1 x sin y) = sin y. 

cos (360° + y) - cos 360° cos y - sin 360° sin y 

— (1 x cos ?/) — (0 x sin 2/) = cos y. 

tan (360° + y) = ^^ = tan y. 
v 7 cosy 

cot (360° + y)= ^^- = cot y. 
sin ?/ 

10. Find, by Formulas [4]-[ll], the first four functions of x — 90°. 

sin (x — 90°) = sin x cos 90° — cos x sin 90° 

= (0 x sin x) — (1 x cos x) = — cos x. 

cos (x — 90°) = cos x cos 90° 4- sin x sin 90° 

= (0 x cos x) + (1 x sin x) = sin x. 

tan (x — 90°) = = — cot x. 

sinx 

sin t 
cot (x - 90°) = = - tan x. 



11. Find, by Formulas [4]-[ll], the first four functions of x - 180° 

sin (x - 180°) = sin x cos 180° - cos x sin 180° 

= sin x ( — 1) — cos x x = — sin x. 

cos (x - 180°) = cos x cos 180° + sin x sin 180° 

= cos x ( — 1) + sin x x = — cos x. 

tan (x - 180°) = _LL± - tan x. 

— COS X 

cot (x - 180°) = ~ C0S X = cot x. 



teachers' edition. 77 

12. Find, by Formulas [4]— [11]. the first four functions of x - 270°. 

sin (x — 270°) = sin x cos 270° — cos x sin 270° 

= sin x x — cos x x ( — 1) = cos x. 

cos (x — 270°) = cos x cos 270° + sin x sin 270° 

= cos x x + sin x ( — 1) = — sin x. 

tan (x — 2 /0°) = = — cot x. 

— sin x 

sin x 

cot (x — 270°) = - — — tan x. 

v cos X 

13. Find, by Formulas [4]-[ll]. the first four functions of — y. 

sin (0° — y) = sin 0° cos y — cos 0° sin y 

= (0x cos y) — (1 x sin y) — — sin y. 

cos (0° — y) = cos 0° cos y 4- sin 0° sin y 

= (lx cos y) + (0 x sin 2/) = cos ?/. 

tan (0° - y) = ~ sm2/ = - tan y. 
cos 2/ 

COS ?/ 

cot (0° - y) = — = - cot y. 

— sin ?/ 

14. Find, by Formulas [4]-[ll], the first four functions of 45° — y. 

sin (45° — y) — sin 45° cos y — cos 45° sin y 

— £ V2 cos y — i V2 sin ?/ = $ ^2 (cos ?/ - sin y). 
cos (45° — y) = cos 45° cos ?/ -f sin 45° sin y 

— \ V2 cos y + i V2 sin ?/ = J V2 (cos 2/ + sin ?/). 
cos ?/ — sin y 1 — tan y 



tan (45° - ?/) 
cot (45° - y) = 



cos ?/ 4- sin ?/ 1 + tan ?/ 
cos ?/ -f sin y _ cot y + 1 
cos ?/ — sin 2/ cot 2/ — 1 



15. Find, by Formulas [4]-[ll], the first four functions of 45° + y. 

sin (45° + y) = sin 45° cos y + cos 45° sin y 

= i V2 cos y + i V2 sin y = -J- V2 (cos 2/ + sin y). 
cos (45° + y) = cos 45° cos y — sin 45° sin y 

= i V2 cos y — i V2 sin y = -J- V2 (cos 2/ — sin y). 

tan (45° + y) = C0Sy + siny = 1+tany . 
cos 2/ — sin ?/ 1 — tan y 

cos 2/ — sin y cot 2/ — 1 



cot (45° + y) = 



cos ?/ + sin 2/ cot 2/ + 1 



78 



PLANE TRIGONOMETRY. 






16. Find, by Formulas [4]-[ll], the first four functions of 30° -f y. 
sin (30° + y) — sin 30° cos y -f cos 30° sin y = -J (cos y + Vssiny). 
cos (30° 4- y) = cos 30° cos y — sin 30° sin y = i ( V3 cos y ~ sin y). 

tan (30° + y) = C °l y — ^ ; divide each term by V3 cos y, 

v3 cos y — sin y 

i Vs 4- tan i/ 
1 — -J- V3 tan ?/ 

■yQ cos 2/ sin ?/ 

cot (30° 4- y) = ; divide each term by sin y, 

cos y + V3 sin y 

_ V3 cot y - 1 
cot y + V3 

17. Find, by Formulas [4]-[ll], the first four functions of 60° — y. 
sin (60° — y) = sin 60° cos y — cos 60° sin y = -J- ( V3 cos y — sin y). 
cos (60° — y) = cos 60° cos 2/ 4- sin 60° sin y = -J (cos 2/ 4- V3 sin y). 

tan(60O-2/) = V ^ C ° S2/ ~ Sin2/ - ^"^ 



cot (60° - y) = 



cos ?/ 4- V3 sin 2/ 14- ^3 tan 2/ 

cos 2/ 4- ^3 sin y _ i "v^ cot 2/4-1 

V3 cos 2/ — sin 2/ cot y — i V3 



18. Find sin 3 x in terms of sin x. 
sin 3 x = sin (2 x 4- #) 

= sin 2 # cos x 4- cos 2 # sin x. 
sin 2 x = 2 sin x cos x. 
cos 2 x = cos 2 x — sin 2 x. 
.*. sin 3 x 

= 2sinx cos 2 x 

4- sin x cos 2 x — sin 3 x 
= 3 sin x cos 2 x — sin 3 x. 

But cos 2 x = 1 — sin 2 x. 
.-. sin 3 x 

= 3 sin x — 3 sin 3 x — sin 3 x 

= 3sinx — 4sin 3 x. 

19. Find cos 3 x in terms of cos x. 
cos 3 x = cos (2 x + x) 

= cos 2 x cos x 

— sin 2 x sin x. 



sin 2 x — 2 sin x cos x. 
cos 2 x = cos 2 x — sin 2 x. 
.-. cos3x 

— cos 3 x — sin 2 x cos x 

— 2 sin 2 x cosx 
= cos 3 x — 3 sin 2 x cos x. 
But sin 2 x = 1 — cos 2 x. 
.*. cos3x 

= cos 3 x — 3 cos x 4- 3 cos 3 x 
== 4 cos 3 x — 3 cosx. 

20. Given tan \ x = 1 ; find cos x. 
tan i x : 



VI — cos x 

1 4- cos x 

VI — cos X 
1 4- cos x 
1 — cos X 



1 4- cos x 



TEACHERS' EDITION. 



79 



1 + COS x = 1 — COS X. 

2 cos x = 0. 
cos x = 0. 



21. 

sin x. 



Given cot -J- x = V3 ; find 



coti» = -^ — 



3 = 



cosx 
1 + cos x 









1 - 


cosx 


-3 


cosx 


= 


1 + 


cosx. 


-4 


cosx 


= 


-2. 






cosx 


= 


1 
2* 






sin 2 x 


r= 


1 - 


COS 2 X 






= 


1 - 


1_ 3 
4~ 4 



4-^, 



22. Given sin x = 0.2 ; find sin £ x 
and cos \ x. 

sinx = 0.2. 
cos 2 x = 1 — sin 2 x 
= 1 - 0.04. 
cos x = VCK96. 



sin-J-x 



-4- 



— COSX 



=^ 



=aM 



.4 Ve 



= 0.10051. 



cos|x 



-4 



4- cosx 



=4^ 



4V6 



= 0.99493. 



23. Given cos x = 0.5 ; find cos 2 x and tan 2 x. 
cos 2x = c os 2 x — si n 2 x. 
sin x = Vl -0.5 2 = \ Vs. 
.-. cos2x =0.25 -0.75 
1 

2' 

iV3 



= _0.50=- 



tan x = 



tan 2 x 



sin x 
cosx 

2 tan x 
1 - tan 2 x 1-3 



■ = V8 

2 V3 



= -V3. 



24. Given tan 45° = 1 ; find the functions of 22° 30'. 
Let x = 45°. 

sin x 



tan x = 



= 1. 



cosx 
. sin x = cos x. 
sin 2 x + cos 2 x = 1. 
2 sin 2 x = 1. 
sin 2 x = \. 
sin x = £ V2 : 



80 



PLANE TRIGONOMETRY. 



sin 22° 30' = sin £ x 



« 22° 30' = cos i x = ^y 1 + ^" 2 = i V2 + V2 = 0. 

} Q30^tani^\/^i^^A/ (1 -^ V ^ =: V2(l-iV2) 2 
1 + i V2 * 1 - * 



3827. 
9239. 



tan 22 



cot 22 



= (1 - 1 V2) V2 = V2 - 1 .= 0.4142. 

:\2 



! °30^cotix = A/ 1+ ^ V LV (1+ ^ ) LV2(l +i V2) 2 
■1 - £ V2 * 1 - i 



: (1 + i V2) V2 = V2 + 1 = 2.4142. 



25. Given sin 30° = 0.5 ; find the functions of 15°. 
sin 30° = 0.5 = -. 



,.cos30o = X /T^ = ^ = i V3. 



r*=\- 



— cosx 



.-. sin 15° = 



P = A /iz±^- t vjr 



V3 = 0.2588. 



cos 15° = \ * + "^ 3 = i V2TV3 = 0.9659. 



tan 15° 



/ l -j V3 _ / 2 - V3 
\ ' 1 + i V3 \ 2 + V3 



= V( 2 - V ^ = 2-V3 = 0.267^ 
* 4-3 



cot 15° 



S 



■V3 



iV3 



rr 2 + VS = 3.7321. 



26. Prove that 

. sin 33° + sin 3° 

tan 18° = 

cos 33° + cos 3° 

Let x = 18°, 
and y = 15°. 

Then 
(1) 2 sin x cos y 

= sin (x-f- ?/) + sin (x-?/). 



(2) 2 cos x cos ?/ 

= cos(x+2/) + cos(x— 2/). 
Divide (1) by (2), 

tan x = sin fo + y)+shi(x-2/) 
cos(x+2/) + cos(x— y) 
Substitute values of x and y, 
sin 33° + sin 3° 



tan 18° 



' cos 33° + cos 3° 



TEACHERS EDITION. 



81 



27. Prove the formula 
2 tanx 



sin 2 x 


l + tan 2 x 


sin 2 x 


= 2 sin x cos x 




2 sin x 

= cos 2 X 

cosx 




2 sin x 1 

= x — z~ 

cosx sec 2 x 


But by Prob. 2, Ex. V, 


sec 2 x 


= 1 4- tan 2 x. 


. sin 2 x 


2 tanx 



1 -f tan 2 x 

28. Prove the formula 
1 — tan 2 x 



COS 2 X : 



COS 2 X : 



1 + tan 2 x 
sin 2x 
tan2x 

2 sin x cos x 

2 tanx 
1 — tan 2 x 
(l-tan 2 x)(2sinx cosx) 



2 tanx 

= (1 — tan 2 x) cos 2 x. 
But by Prob. 2, Ex. V, 
sec 2 x = 1 + tan 2 x : 



1 



cos^x. 



cos 2 x = 



1 — tan 2 x 
1 + tan 2 x 



29. Prove the formula 
sinx 



tan $ x : 



1 4- cos x 



tan \x 



-4 



— cosx 



(1 — cosx) (1 4- cosx) 
(1 4- cos x) 2 
y/\ — COS 2 X 
1 4- cos x 
sinx 



1 + cos x 

30. Prove the formula 
sin x 



cot -J-x = - 



1 — cos x 



cot-^-x 



-V£ 



cosx 
/(I 4- cosx)(l — cosx) 



(1 — cosx) 2 



VI 



1 — COS X 

sinx 
1 — cos x 



31. Prove the formula 



sin \ x ± cos J- x = v 1 ± sm x. 

sin |- x ± cos | x = V(sin ix ± cos-^-x) 2 



= VsinHx 4- cos 2 ix ± 2 siu £x cos 4-x 
= V 1 -t- 2 sin -J- x cos £ x 



32. Prove the formula 
tan x ± tan ?/ 

cot X ± cot ?J 



: Vl ± si 



± tan x tan y. 



82 PLANE TRIGONOMETRY. 



sin x sin y 

. . ± ± 

tan x ± tan y cos x cos y 



cot x ± cot y cosx cos?/ 
sin x sin y 



sin x cos y ± cos x sin y sin x sin y 



cos x cos ?/ cos x sin ?/ ± sin x cos y 

sin x sin ?v 

= ± = ± tan x tan y. 

cosx cosy 



33. Prove the formula 

1 — tan x 



tan (45° - x) = 
tan (45° - x) 



1 + tan x 
tan 45° — tan x 1 — tan x 



1 + tan 45° tan x 1 -f- tan x 



34. If A, B, C are the angles of a triangle, prove that 
sin A + sin B 4- sin C ■= 4 cos -J- .A cos -J- # cos i C. 
sin J. + sin B -f sin G 

= sin .4 + sin 5 + sin [180° - (A + #)] 

= sin ^4 + sin I? + sin (A 4- B) 
By [20] and [12], 

= 2sini(^L + B)cosi(A - B) 4- 2smi(A + B)cosi(A + B) 

= 2 sini (-4. + B) [cos* (-4 - B) + cos£ (J. + £)] 
By [22], =2sini(A + B)2cosiAcosiB 

= 4sini (A -{- B) cos £ A cos £ 5. 
But cos i C = cos [90° -i(A + B)]= sin i (4 + #). 

.-. sin A -f sin I? 4- sin C = 4 cos | J. cos i B cos £ C 

35. If A , J3, C are the angles of a triangle, prove that 

cos A 4- cos J? 4- cos C = 1 4- 4 sin £ ^4 sin ^ i? sin -} C. 
cos C = cos [180° -(A + B)]=- cos (^4 4- 5). 
.-. cos A 4- cos B 4- cos C 

= cos A 4- cos 5 — cos (^4 4- B) 
By [22], = 2 cos \(A 4- #) cos -J- (^4 - 5) - cos (A 4- £) 
By [13], = 2 cos£(^4 4- #) cos \(A - B) - 2 cosH(^ 4- #) 4- 1 

=r[2cosi(^4 4-^)][cosi(^L -£)-cos£(^4 + £)] + 1 
By [23], = [2 cos i (J. + B)] x [2 sin ± A sin £ 5] 4- 1 

= (2 sin i C) (2 sin \ A sin£E) 4- 1 

= 1 4- 4 sin i A sin i B sin J C. 



TEACHERS EDITION. 



*3 



36. If A. B. C are the angles of a triangle, prove that 

tan A + tan B + tan C = tan A x tan B x tan C. 
Since A + B + C= 180 c . C = 180"- - (A + 5). 

.-. tanC = tan [180" - (-4 4- £)] = -tan(J. - B). 
By [6], tan J. + tan B = tan (.4 - B) (1 - tan J. tan B) 

= tan (A + B) — 12lti(A-tB) tan .4 tan B. 
.-. tan ^4 -f tan B + tan C = tan (^1 — I?) — tan (J. — B) 

- tan (.4 + B) tan ^1 tan B 
= - tan (J. + B) tan J. tan B 
= tan J. tan B tan C. 

37. If A, B, C are the angles of a triangle, prove that 

cot } ^ 4- cot \ B 4- cot | C = cot 4- A x cot J 5 x cot J C. 

Since M+*B + iC= ^ r,: - *C = 90°-i(i4 +£). 

.-. cotiC = tan | (J. -f 5). 
cot^JB = tan ^M - Ci. 
and cot 4 A = tan } (# - f 

.-. cot i A -f cot ± 2? 4- cot } C 

= tan | (.4 + B) 4- tan } (4 + C) 4- tan } (B - I i 
By Prr.b. 36, = tan \ {A 4- B) x tan ± (J. 4- C) x tan *(B - 

= cot\A ■ cot I 7; • col I C. 



38. Change cot x 4- tan x to a 
form more convenient for logarith- 
mic computation. 

cot x 4- tan x 

cos x sin x 

= -s + 



By [12], 



sin x cos x 
9<*x-|- sin-x) 

2 sin x cos x 
2 



sin 2 x 



39. Change cot x — tan x to a 
form more convenient for logarith- 
mic computation 

cot x — tan x 

cos x sin x 



sin x cos x 



By [13]. = 



By [12], = 



cos 2 x — sin-x 
sin x cos x 
cos2x 



sm x cos x 
2 cos2x 
2 sin x cos x 
2 cos 2 x 



sin 2 x 
= 2 cot 2 x. 

40. Change cot x — tan y to a 
form more convenient for logarith- 
mic computation. 

cot x - 1 - tan y 

sin y 
~~ sin x cos s 

cos x cos y — sin x sin y 



By [9], = 



sin x cos y 
cos(x - y) 
sin x cos y 



84 



PLANE TRIGONOMETRY. 



41. Change cot x — tan y to a 
form more convenient for logarith- 
mic computation. 

cot x — tan y 

_ cos x sin y 
~ sin x cos y 
_ cos x cos 2/ — sin x sin ?/ 
sin x cos y 

^cosfr + y), 

L J since cos 2/ 

>. « ^ 1 — cos 2 x ^ , 

42. Change to a form 

1 -f cos2x 

more convenient for logarithmic 
computation. 

1 — cos 2 x 
1 — cos 2 x 2 



1 -1- cos 2 x 1 + cos 2 x 



By [16], [17], = 



COS 2 X 

tan 2 x. 



43. Change 1 + tan x tan y to a 
form more convenient for logarith- 
mic computation. 
1 + tan x tan 2/ 

sin x sin 2/ 
■ = 1 h x - 

COS X COS 2/ 

_ cos x cos ?/ + sin x sin ?/ 



By [9], 



cos x cos y 
cos (x — i/) 
" cos x cos 2/ ' 



44. Change 1 — tan x tan y to a 
form more convenient for logarith- 
mic computation. 

1 — tan x tan y 

sin x sin y 
cos x cos y 
__ cos x cos ?/ — sin x sin ?/ 
cos x cos y 



By [5],= 



cos (x + ?/) 
cos x cos y ' 



45. Change cot x cot y + 1 to a 
form more convenient for logarith- 
mic computation. 

cot x cot 2/ + 1 

cos x cos y 
= - — X - — - + 1 

sin x sin y 

_ cos x cos 2/ + sin x sin ?/ 
sin x sin y 

By[0], = c ? a(iC T y) - 

sin x sin 2/ 

46. Change cot x cot 2/ — 1 to a 
form more convenient for logarith- 
mic computation. 

cot x cot y — 1 
_ cos x cos y 
~ sin x sin y 
cos x cos 2/ — sin x sin 2/ 



By [6],= 



sin x sin ?/ 
cos (x + y) 
sin x sin 2/ 



47. Change 



tan x + tan 2/ 



to a 



cot x + cot y 
form more convenient for logarith- 
mic computation. 

tan x + tan y 

cot x + cot 2/ 

sin x sin v 

+ 

_ cos x cos y 

~ cos x cos 2/ 

sin x sin y 

sin x cos 2/ + cos x sin 2/ 
_ cos x cos 2/ 

"" sin x cos 2/ + cos x sin 2/ 

sin x sin 2/ 
_ sin x sin y 
~ cos x cos y 
= tan x tan 2/. 



TEACHERS EDITION. 



85 



Exercise XV. Page 63. 

1. Find all the values of the following functions: sin-^VS, 
tan-^VS, vers— H, cos- 1 (— £ V2), csc- 1 V2, tan- 1 go, sec -1 2, 
cos-i(-*V3). 

sin- 1 i V3 = 60° + 2 nit or 120° + 2 nit. 

twr^Vs = 30° + 2 nit or 210° + 2n7r. 

vers- ! i = 60° + 2 ntf or 300° + 2 nit. 

cos- i ( - 1 V2) = 135° -f 2 w*r or 225° + 2 nit. 

esc- 1 V2 = 45° + 2 nit or 135° + 2 Bar. 

tan- 1 ao = 90° + 2 nit or 270° -f 2 utt. 

sec- 1 2 = 60° + 2 nit or 300° + 2 nit. 

cos -i (_ j. V3) = 150° + 2 riTT or 210° + 2 titt. 

2. Prove that sin— Y ( — x) —— sin— J x ; cos - x ( — x) = it — cos - ] x. 

sin -1 (— x) = the angle whose sine is — x 
= — the angle whose sine is x 
= — sin - l x. 

cos- 1 ( — x) = the angle whose cosine is — x 
= 7r — the angle whose cosine is x 

= 7T — COS - 1 X. 



3. If sin- 1 x + sin- 1 ?/ = it, prove 
that x — y. 

If the sum of two angles is 180°, 
the sine of the one is equal to that 
of the other. 

Hence, sin (sin— x x) = sin (sin - 1 y). 
.-. x = y. 

4. If y = sin— l i, find tan y. 
2/ = sin— 1 |. 

•'• sin y = J 

cos y = Vl — sin 2 ?/ = V| : 



.V5. 



tan y = — 5— = — L_ = i V2. 

f.vS 2V2 

5. Prove that 

cos (sin- 1 x) = Vl — x 2 . 
Let sin— l x = y ; 



then 



vr 



x = sm 2/, 
x 2 = cos 2/ 

= cos (sin - 1 x). 



6. Prove that 
cos (2 sin- ix) = 1 -2x 2 . 
Let sin- l x = y ; 
then x == sin 2/, 

cos (2 sin— x x) = cos 2 y 

— cos 2 2/ — sin 2 2/ 
= 1 — sin 2 2/ — sin' 2 y 







= 


1 


— 2 sin 2 2/ 






= 


1 


-2x 2 . 


7. Prove that 
tan (tan - 1 x -f- tan 
Let tan— x x = 


-] 
14, 


1 -xy 


and. 


tan- 


-ly = 


U 


a 



86 



PLANE TRIGONOMETRY. 



then x = tan u, 

y = tan v, 
tan (tan- : x -f tan~ l y) 
= tan (u + v) 
tan u -f tan v 
1 — tan u tan v 
_ x + y 
I -xy 

8. If x = VI, find all the values 
of sin— l x + cos— l x. 

sin- 1 V| = 45°-f-2n7r or 135° + 2ft7T. 
cos-iVi =45° + 27i7ror-45°+2n7r. 



.\ sin" 



-iVT-H 



-iVi 



= 90° + 2 titt, 2 ntf. 

180° + 2 ntf, or 90° -\-2mt 
= 0°, 90°, or 180°. 



9. Prove that 
x 



tan -1 



( , ) = sir 



Let 
tan 

then 



-( 



vr 



=) = *; 

2/ 

; = tan y. 



Vl-x 2 
By Prob. 2, Ex. V, 

sec 2 ?/ = 1 4- tan 2 y 

1 



VT~~ 



1 -X 2 

1 



.-. tan -1 



sec 2 y 

: 1 - 

. sin y = x. 
.-. ?/ = sin - 

V Vl -x 2/ 



1 -x 2 . 



10. Find the value of 

sin (tan- ij 5 ^). 
Let tan- 1 T \ = x ; 
then T 5 2 = tan x. 

By Prob. 2, Ex. V, 

sec 2 x = 1 + tan 2 x 

= i + (A) 2 

— 169 

— T4?* 

.'. cos 2 x = iff. 
.-. sin 2 x = T 2^V 
.-. sin x = ± T 5 g. 

11. Find the value of 

cot (2 sin- 1 !). 
Let 
then 



sin -1 ! = x ; 



sin x = !, 
cos x = ± f , 



cot (2 sin- 1 !) = cot2x 

_. rK _ cot 2 x - 1 

By [51, = 

J L J 2cotx 

V -1 



= ± 



7 



24 



12. Find the value of 

sin (tan— 1 J + tan -1 J). 
Let tan— x | = x, 





tan- 1 1 = 2/; 


then 


tan £C = J, 


and 


tan y '= fc 


Now 


sec 2 x — 1 + tan 2 x 




_ 5 
— 4> 




cos 2 x = f , 




±2 
cos X = — — » 

V5 




sin x = ; 

V5 


and 


i3 
cos y = — — » 



Vio 



TEACHERS EDITION. 



87 



±1 

sin y = — — : • 
VK) 
,\ sin (tan— !i + tan— x |) 
= sin (x + y) 

= sin x cos y -f cos x sin y 
1 3 2 1 

V6 Vio Vs VTo 



V50 
= ±*V2. 



13. If sin-ix = 2 cos~ *x, find x. 

sin-ix = 90° — cos _1 x. 
, 90° — cos - 1 x = 2 cos - : x, 
3cos- 1 x = 90°, 
cos-!x = 30°, 150°, or 270°, 
x = ± ' J V3, or 0. 



14. Prove that 
tan (2 tan— x x) ; 



2x 
1 -x 2 ' 



Let tan— l x = y ; 
then x = tan ?/, 

tan (2 tan— l x) = tan 2 y 
2 tan ?/ 









1 -tan 2 ?/ 








2x 

1 -X 2 


15. 


Prove that 


sin (2 
Let tan- 


tan - 

-ix 


. zx 
; 1 +x 2 

= y\ 


then 




X 


= tan ?/, 


sin (2 


tan~ 


■!X) 


= sin 2 y 

= 2 sin ?/ cos ?/ 

= 2 S ^lcos^y 
cosy 

2 tan ?/ 

sec 2 y 

2ta,ny 
1 -f tan 2 ?/ 

2x 



1 + x 2 



Exercise XVL Page 67, 



1. What do the formulas of Sect. 
XXXIV, p. 64, become when one of 
the angles is a right angle ? 




If angle C is a right angle, 
a _ sin A 
c sin C 



= sin ^4 



sin B 



= sin B ; 

— tan A ; 
= c; 



sin B sin C 

2. Prove by means of the Law of 
Sines that the bisector of an angle 
of a triangle divides the opposite 
side into parts proportional to the 
adjacent sides. 



c 




sin 


C 


a 
b 


= 


sin 
sin 


A 
B 


a 


= 


c 




sin^l 


sin 


C 


b 




c 





88 



PLANE TRIGONOMETRY. 



Let CD bisect 


angle C ; 




AD 


sin 


iC 




CD 


sin A 


id 


DB 


sin 


\c 




CD 


sin B 


By division, 








AD 


sin 


B 




DB 


sin 


A 


But 


sin B 

sin A 


__b 
a 






AD 


b 






" DB 


a 





3. What does Formula [26] be- 
come when A = 90° ? when A = 0° ? 
when A = 180° ? What does the tri- 
angle become in each of these cases ? 

Formula [26] is 

a 2 = b' 2 + c 2 - 2 be cos A. 

1. When A = 90°, cosA = 0. 
.-. a 2 = b 2 4- c 2 . 

2. When A = 0°, cos A = 1. 
.-. a 2 = 6 2 + c 2 - 2 6c. 

3. When A = 180°, cos A = - 1. 
.-. a 2 = 6 2 + c 2 + 2 6c. 

4. A right triangle. 

5. A straight line. 

A ? C 



6 =AC. 
6. A straight line. 
A 



c = AB. 

a = b — c. 



B 

a = BC. 
b =AC. 



c = BA. 
a = b + c. 



4. Prove (Figs. 56 and 57) that 
whether the angle B is acute or 
obtuse c = a cos 2? + 6 cos A. What 
are the two symmetrical formulas 



obtained by changing the letters? 
What does the formula become 
when jB = 90°? 




A 


C £ D 




Fig. 2. 


Case I. 


When angle B is acute 


(Fig. 1). 


cos 5 == 

a 




AD 



cos A = 

.-. DB = a cos B, 
and AD = 6 cos A. 

.-. DB+AD = acos 2?+ 6 cos A. 
But DB-\-AD = c. 

.-. c = a cos 2? + 6 cos A. 

Case II. When angle 5 is obtuse 
(Fig. 2). 

AD 

6 

BD 

a 

= — cos B. 
.\ AD = b cos A, 
and jBD = — a cos J5. 



cos A. 
= cos (180° - B) 



TEACHERS' EDITION. 



89 



.-. AD — BD = bcosA+acosB. 
But AD -BD = c. 

.*. c = a cos I? +& cos -4. 
The symmetrical formulas are 

b — acosC-f ccos^4, 
and a = b cos C + c cos B. 

When B = 90°, 

A C 

cos A — - • 
b 

.-. c = bcosA. 

5. From the three following equa- 
tions (found in the last example) 
prove the theorem of Sect. XXXV, 
p. 66 : 

c = a cos I? + b cos A, 
b = a cos C + c cos A, 
a = b cos C + c cos 5. 
c 2 = ac cos 5 -f 6c cos A. (1) 
6 2 = a&cos C -f 6c cos JL (2) 
a 2 = a6 cos -f «c cos 2?. (3) 
Add (2) and (3), 
a 2 + 6 2 = 2 a6 cos C + be cos ^. 

-I- ac cos B. (4) 

Subtract (4) from (1), 

c 2 — a 2 — 6 2 — — 2 ab cos C. 
.-. c 2 = a 2 + 6 2 - 2 a& cos C. 
In a similar manner it may be 
proved that 

a 2 = 6 2 + c 2 -2 6c cos -4, 
and 6 2 = a 2 + c 2 — 2 ac cos B. 

6. In Formula [27] what is the 
maximum value of -j- (-4 — jB) ? 

a — 6 __ tan i ( J. — 2?) 

a + 6 ~ tan \ {A + 5) 
The limit of A - B is 180°. 
Therefore, the limit of the maxi- 
mum value of i (A - B) 

= 1^ = 90°. 

2 



7. Find the form to which For- 
mula [27] reduces, and describe the 
nature of the triangle, when 

(i) C = 90 ; 

(ii) A - B = 90°, and B = C. 
a - 6 _ tan \ (A . - B) 
a + 6 ~ tan \ {A + J5) ' 

(i) When C = 90°. 

J. + B = 90°. 

5 = 90° -.4. 

a-b _ tan jj- [^1 - (90° - A)] 
a + 6 tan 45° 

_ tan (A - 45°) 
1 

= tan (4 -45°). 

Since C is a right angle, the tri- 
angle is a right triangle. 

(ii) When A-B=90°, and B=C. 

a-b _ tan .V (^L - 5) 
a + 6 ~ tan | (vl + 2?) 

A+B+C= 180°, 



or 


J+25 = 180° 




A-B =90° 




.-. 35= 90°, 




5= 30°, 




C = 30°, 


and 


A = 120°. 




a — 6 tan 45° 




a -f 6 tan 75° 




tan 45° 




cot 15° 




1 



2 +V3 
... a + 6 = (a - 6) (2 + V3). 

Since B — C, the triangle is isos- 
celes. 



90 



PliANE TRIGONOMETRY. 



Exercise XVII. Page 69. 



1. 


Given find 


3. 


Given find 


a : 


= 500, C = 123° 12', 


a : 


= 804, C = 35° 4', 


A: 


= 10° 12', 6 = 2051.5, 


A 


= 99° 55', 6 = 577.31, 


B: 


= 46° 36' ; c = 2362.6. 


B 


= 45°1'; c = 468.93. 




a = 500. 




a = 804. 




A = 10° 12' 




A= 99° 55' 




£ = 46° 36' 




£= 45° V 




A + B= 56°48 / 




A + B = 144° 56' 




.-. = 123° 12'. 




.-. C = 35° 4'. 




log a = 2.69697 




log a = 2.90526 




cologsin A = 0.75182 




cologsin^. = 0.00654 




log sin £ = 9.86128 




log sin B = 9.84961 




log 6 = 3.31207 




log 6 = 2.76141 




6 = 2051.5. 




6 = 577.31. 




log a = 2.69897 




log a = 2.90526 




cologsin A = 0.75182 




cologsin .4 = 0.00654 




log sin (7 = 9.92260 




log sin (7 = 9.75931 




logc = 3.37339 




logc = 2.67111 




. c = 2362.6. 




c = 468.93. 


2. 


Given find 


4. 


Given find 


a 


= 795, C = 55° 20', 


a 


= 820, C = 25° 12', 


A 


= 79° 59', 6 = 567.69, 


A 


= 12° 49', 6 = 2276.6, 


B 


= 44° 41'; c = 663.99. 


B 


= 141° 59'; c = 1573.9. 




a = 795. 




a = 820. 




A = 79° 59' 




A = 12° 49' 




B= 44° 41' 




B = 141° 59' 




A + B = 124° 40' 




A + B = 154° 48' 




.-. C = 55° 20'. 




.-. C = 25° 12'. 




log a = 2.90037 




log a = 2.91381 




cologsin A = 0.00667 




cologsin A = 0.65398 




logsinJ5 = 9.84707 




log sin 5 = 9.78950 




log 6 = 2.75411 




-| n0 . h — o oc7or) - 








6 = 567.69. 




6 = 2276.6. 




loga = 2.90037 




log a = 2.91381 




cologsin^. = 0.00667 




cologsin^. =0.65398 




log sin C = 9.91512 




log sin C = 9.62918 




logc = 2.82216 




logc = 3.19697 




c = 663.99. 




c = 1573.9. 



TEACHERS EDITION. 



91 



5. 


• Given find 


7. 


Given find 


C : 


= 1005, C = 47° 14', 


a 


= 6412, B = 56° 56', 


A -. 


= 78° 19', a = 1340.6, 


A 


= 70° 55', 6 = 5685.9, 


B-- 


= 54° 27'; 6 = 1113.8. 


C 


= 52° 9'; c = 5357.5. 




c = 1005. 




a = 6412. 




^1 = 78° 19' 




^4 = 70° 55' 




B= 54°27 / 




C= 52° 9' 




.4 + B = 132° 46 / 




A + C = 123° 4' 




.-. C = 47° 14'. 




.-. J5 = 56° 56'. 




log c = 3.00217 




log a = 3.80699 




colog sin C = 0.13423 




colog sin A = 0.02455 




log sin A =9.99091 




log sin B= 9.92326 




log a = 3.12731 




log b = 3.75480 




a= 1340.6. 




b = 5685.9. 




logc = 3.00217 




loga = 3.80699 




cologsin (7 = 0.13423 




colog sin A = 0.02455 




logsinJ3 = 9.91042 




log sin (7 = 9.89742 




log b = 3.04682 




logc = 3.72896 




6 = 1113.8. 




c = 5357.5. 


6. 


Given find 


8. 


Given find 


6: 


= 13.57, A = 108° 50', 


6: 


= 999, B = 77° 


B-. 


= 13° 57', a = 53.276, 


A : 


= 37° 58', a = 630.77, 


C-- 


= 57° 13'; c = 47.324. 


C: 


= 65° 2'; c = 929.48 




6 = 13.57. 




6 = 999. 




J5= 13° 57' 




^ = 37° 58' 




(7= 57° 13' 




(7= 65° 2 / 




£ + C = 71° 10' 




^4 + C = 103° 




.-. A = 108° 50'. 




.-. £= 77°. 




log 6 = 1.13258 




log 6 = 2.99957 




cologsinjB = 0.61785 




colog sin £ = 0.01128 




log sin A = 9.97610 




log sin A =9.78902 




loga = 1.72653 




log a = 2.79987 




a = 53.276. 




a = 630.77. 




log 6 = 1.13258 




log b = 2.99957 




colog sin B = 0.61785 




colog sin B = 0.01128 




log sin C = 9.92465 




log sin C = 9.95739 




logc = 1.67508 




log c = 2.96824 




c = 47.324. 




c = 929.48. 



92 



PLANE TRIGONOMETRY. 



9. In order to determine the dis- 
tance of a hostile fort A from a 
place B, a line BC and the angles 
ABC and BCA were measured, and 
found to be 322.55 yards, 60° 34', 
and 56° 10', respectively. Find the 
distance AB. 

a = 322.55. 
B= 60°34 / 
C = 56° KK 
J5 + C = 116° 44' 
.-. A - 63° 16' 
log a = 2.50860 
cologsin A = 0.04910 
log sin C = 9.91942 
log c = 2.47712 
c = 300. 
.-. AB = 300 yards. 

10. The angles B and C of a tri- 
angle ABC are 50° 30" and 122° 9', 
respectively, and BC is 9 miles. 
Find AB and .40. 



C: 

J5+ O: 

.'.^4 : 

log BC -- 

colog sin ^4 : 

log sin B : 

. log 6 : 

6: 

.'. AC: 

log £C : 

colog sin A : 

log sin C -. 

logc: 

C : 
.-. AB : 



:122° 9 / 

50°30 / 
: 172° 39 ; 

7° 21'. 
: 0.95424 
: 0.89303 
: 9.88741 
: 1.73468 
: 54.285. 
: 54.285 miles. 
: 0.95424 
; 0.89303 
; 9.92771 
: 1.77498 
: 59.564. 
: 59.564 miles. 



11. Two observers 5 miles apart 
on a plain, and facing each other, 
find that the angles of elevation of 
a balloon in the same vertical plane 
with themselves are 55° and 58°, 
respectively. Find the distance 
from the balloon to each observer, 
and also the height of the balloon 
above the plain. 

B= 58° 
A = 55° 
A + B = 113° 
.-. C = 67°. 
logc = 0.69897 
cologsin = 0.03597 
log sin A = 9.91336 
log a = 0.64830 
a = 4.4494. 
.-. BC = 4.4494 miles. 

logc = 0.69897 

cologsin (7 = 0.03597 

log sin B = 9.92842 

log 6 = 0.66336 

b = 4.6064. 

.-.AC = 4.6064 miles. 



I'o find h. 



h 

a ' 



■■ sin B. 



.-. h = a sin B. 

log a = 0.64830 

log sin B = 9.92842 

log h = 0.57672 

h = 3.7733. 

.-. height = 3.7733 miles. 

12. In a parallelogram, given a 
diagonal d and the angles x and y 
which this diagonal makes with the 
sides. Find the sides. Compute 
the results if d = 11.237, x = 19° 1', 
and y = 42° 54 7 . 



TEA< HERS EDITION. 



93 



x -f V : 

.". Z : 
log tZ : 

colog sin 2 : 

log sin x ■■ 

log a: 

a : 

\Ogd: 

colog sin z : 
log sin y ■ 

lOgC : 



smx 
sin z 
dsinx 
sin z 
sin?/ 
sin z 
d sin ?/ 
sin z 
: 11.237. 
19° V 
42° 54 / 
: 61° 55' 

: 118° 5' 

: 1.05065 
; 0.05440 
: 9.51301 
: 0.61806 

: 4.1501. 
: 1.05065 
: 0.05440 

: 9.83297 
: 0.93802 

:8.67. 



13. A lighthouse was observed 
from a ship to bear N. 34° E. ; after 
the ship sailed due south 3 miles, it 
bore N. 23° E. Find the distance 
from the lighthouse to the ship in 
each position. 

c = 3. 
A = 23° 
B = (180° - 34°) = 146^ 
A + B = 169° 
.-. C = 11°. 
log c = 0.47712 
colog sin C = 0.71940 
log sin A = 9.59188 
loga = 0.78840 
a = 6.1433. 



logc = 0.47712 

colog sin C = 0.71940 

log sin B = 9.74756 

log 6 = 0.94408 

6 = 8.7918. 

Therefore, the required distances 

are 6.1433 miles and 8.7918 miles. 

14. In a trapezoid, given the 
parallel sides a and b. and the 
angles x and y at the ends of one 
of the parallel sides. Find the non- 
parallel sides. Compute the results 
when a = 15, b — 7, x — 70°. y = 40°. 

Given parallel sides, 

AB = 7 and DC = 15 ; 

also,^LDC=40° and BCD = 70°; 

required AD and BC. 

Draw AE II BC ; 

then AB = EC (ll a comp. bet. Il 8 ), 

and DE = DC - AB 

= 15 - 7 = 8. 

Also AED=BCD = 10°(ext. int. A). 

Now 

IM^ 180° -(40° + 70°) = 70°. 

But since 

AED = DAE = 70°, 

the A AED is isosceles, and 

DA=DE = 8. 

Now AE = EC, and we are to 

find BC. 

AE _ sin ADE 

1)E~ sin DAE' 

log DE = 0.90309 

log sin ADE = 9.80807 

colog sin D J..E = 0.02701 

log AE = 0.73817 

AE = EC = 5.4723. 

15. Given 6 = 7.07107, A = 30°, 
C = 105° ; find a and c without 
using logarithms. 



94 



PLANE TRIGONOMETRY. 



Let p and q denote the segments 
of c made by the _L dropped from C. 

A = 30°, 

C = 105°. 

.-. B = 45°. 

a _ sin A _ -J- 

b sin J5 i V2 

6 7.07107 c 

a = = = 5. 

V2 1.41421 

| = cos -A = i Vi = 0.86603. 
o 

p = b x 0.86603 

= 7.07107 x 0.86603 

= 6.12376. 

? = cos 5 = £ V2 = 0.70711. 
a 

q = ax 0.70711 

= 5x0.70711 =3.53555. 
c =p + g 

= 6.12376 + 3.53555 

= 9.6593. 

16. Given c = 9.562, ^1 = 45°, 
B — 60° ; find a and b without using 
logarithms. 





= 


75°. 










a = 


csinA 
sin C 








sin 


C = 


sin (45° 


+ 30°) 








= 


sin 45° 


30S 30° 












4- cos 45° 


sin 


30° 



= iV2xiV3-f ^V2xl 
= i(V6+ V2). 
_ 9.562 xjV2 
~i(V6+ V2) 
_ 19.124 x V2 
V6 +V2 
(19.124 x V2)(V6- V2) 



= 9.502 (V3-1) 



6 = 



= 6.99986 = 7. 

a sin B 7x4- V3 



sin ^L j. V2 

_ 7 V3 _ 7 V6 
" V2 = 2 

= 3.5 V6 = 8.573. 



17. The base of a triangle is 600 
feet, and the angles at the base are 
30° and 120°. Find the other sides 
and the altitude without using log- 
arithms. 

AB = 600. 

A = 30°. 

B = 120°. 

.-. C = 30°. 

.:a = c = 600 feet. 

a sin B 
b = 



sin A 
_ 600 x sin (180° - 60°) 

sin 30° 
_ 600 x J V3 

= 600x 1.732051 
= 1039.2. 
h = a sin B = 600 x i V3 
= 519.6 feet. 

18. Two angles of a triangle are, 
the one 20°, the other 40°. Find 
the ratio of the opposite sides with- 
out using logarithms. 

Let x = 20°, 

y = 40°, 
and a and b be opposite sides, 
sin x _ a 
siny b 



Then 



TEACHEKS EDITION. 



95 



nat sin x = 0.3420. 
nat sin?/ = 0.6428. 
.-. a : b = 3420 : 6428 
= 855 : 1607. 

19. The angles of a triangle are 
as 5 : 10 : 21, and the side opposite 
the smallest angle is 3. Find the 
other sides without using logarithms. 

Since the angles A, B, C are as 
5:10:21, 

A = jft> of 180° = 25°. 
B = if of 180° = 50°. 
G = H of 180° = 105°. 
, a sin B 3x0.7600 



sin A 

5.438. 



0.4226 



asinC 3x0.9659 



sin A 

= 6.857. 



0.4226 



20. Given one side of a triangle 
equal to 27, the adjacent angles 
equal each to 30° ; find the radius 
of the circumscribed circle without 
using logarithms. 



2E 



sin A 
sin A - sin 120° 
= sin (180° 
sin 60° = \ VS. 

54 .= 18V3. 



-120°)= sin 60°. 



.:2R = 

.-. R = 



\ V3 V3 
9 V3 = 15.588. 



Exercise XVIII. Page 74. 



1. Find the number of solutions 
of the following : 

(i) a = 80, b = 100, A = 30°. 
.-. a<b, 
but a > b sin A = 100 x \ , 

and A < 90°. 

.*. two solutions. 

(ii) a = 50, b = 100, A = 30°. 
.-. a — b sin A = 100 x £. 
.-. one solution. 

(iii) a = 40, b = 100, J. = 30°. 

.*. a < 6 sin ^4 = 100 x J, 
and A < 90°. 

.-.no solution. 

(iv) a=13.4, b= 11.46, A = 77° 20'. 

.-. a > b, 
.-. one solution, 
(v) a = 70, b =z 75, 4 = 60°. 

.-. a < 6, 



but a > b sin ^1 = 75 x i V3, 

and A < 90°. 

.-. two solutions. 

(vi) a= 134.16, 6 = 84.54, 

B= 52° 9' 11". 

6<a, 
£ < 90°, 
nat sin B = 0.7896. 

84.54 < 134.16 x 0.7896. 
.-. b < a sin B. 
.-. no solution. 

(vii) a = 200, b = 100, A = 30°. 
a>6. 
.-. one solution. 



2. Given 

a = 840, 
b = 485, 
A = 21° 31' ; 



find 
1? = 12° 13' 34" 
C = 146° 15' 26" 
c = 1272.1, 



96 



PLANE TRIGONOMETRY. 





colog a = 7.07572 - 10 




loga = 1.95933 




log b = 2.68574 




log sin = 9.99963 




log sin A = 9.56440 




colog sin A = 0.10857 




log sin B = 9.32586 




log c = 2.06753 




B - 12° 13' 34". 




c = 116.82. 




.-. = 146° 15' 26". 








log a = 2.92428 


5. 


Given find 




log sin = 9.74466 


a 


= 55.55, .4 = 54° 31' 13 




colog sin .4 = 0.43560 


b 


= 66.66, = 47° 44' 7 




logc = 3.10454 


B 


= 77° 44' 40"; c = 50.481. 




e = 1272.1. 




log a = 1.74468 
log sin £ = 9.98999 
colog 6 = 8.17613- 10 


3. 


Given find 




log sin A =9.91080 


a 


= 9.399, B = 57° 23' 40", 




A = 54° 31' 13". 


b 


= 9.197, = 2° r20", 




.-. = 47° 44' 7". 


A 


= 120° 35'; c = 0.38525. 




loga = 1.74468 




cologa = 9.02692 - 10 




log sin = 9.86925 




log b = 0.96365 




colog sin A = 0.08920 




log sin A = 9.93495 




logc = 1.70313 




log sin B = 9.92552 




c = 50.481. 




B = 57° 23 / 40". 








.-. C = 2° V 20". 


6. 


Given 




log a = 0.97308 


a = 


309, 6 = 360, A =21° 14' 25' 




log sin = 8.54761 


find 


B= 24° 57' 54", 




colog sin ^4 = 0.06505 




0= 133° 47' 41", 




log c = 9.58574 - 10 




c = 615.67, 




c = 0.38525. 




B' = 155° 2' 6", 
C = 3° 43' 29", 
c' = 55.41. 


4. 


Given find 


There are two solutions, 


a 


= 91.06, J5 = 41°13 / , 


for 


a<b, 


b 


= 77.04, = 87° 37' 54", 


but 


a > 6 sin ^4, 


A 


= 51° 9' 6"; c= 116.82. 


and 


A < 90°. 




colog a = 8.04067 - 10 




log b = 2.55630 




log b = 1.88672 




log sin A = 9.55904 




log sin A = 9.89143 




colog a = 7.51004 - 10 




logsin JB = 9.81882 




log sin B- 9.62538 




B = 41° 13'. 




B= 24° 57' 54". 




.-. = 87° 37' 54". 




,% C = 133° 47' 41". 






TEACHERS EDITION. 



97 



log a = 2.48996 

log sin = 9.85843 

colog sin A = 0.44096 

lose = 2.78935 



c ■■ 



; 615.67. 



Second Solution. 

B' = 180° - B = 155° 2' 6". 

C' = B-A = 3° 43' 29". 

log a = 2.48996 

logsinC" = 8.81267 

colog sin A = 0.44096 

logc' = 1.74359 

c' = 55.41. 

7. Given a = 8.716, b = 9.7* 
A = 38° 14' 12" ; 
find B= 44° 1'28", 

C= 97° 44' 20", 
c = 13.954, 
& = 135° 58' 32", 
C = 5° 47' 16", 
c' = 1.4202. 

There are two solutions, for 
-4 < 90°, a < 6, and > 6 sin ^4 . 

colog a = 9.05968 - 10 
log 6 = 0.99065 
log sin A = 9.79163 
log sin £ = 9.84196 

£= 44° I'M". 
.-. JF = 135° 58' 32". 

.-. C = 97° 44' 20". 
.-. C" = 5° 47' 16". 

log a = 0.94032 

log sin C = 9.99602 

colog sin A = 0.20837 

logc = 1.14471 

C = 13.954. 



log a = 0.94032 

log sin C = 9.00365 

colog sin A = 0.20837 

logc' = 0.15234 

& = 1.4202. 



8. Given 
a = 4.4, 
b = 5.21, 

A = 57°37 / 17 // 

log sin A ■ 

log 5: 

colog a 

log sin 5 : 

B 

lOg 6: 

log COS J. : 
lOgC: 

C : 



find 
B = 90°, 
C = 32° 22' 43", 
c = 2.7901. 

: 9.92661 

: 0.71684 

: 0.35655 - 10 

: 10.00000 

:90°. 

: 32° 22' 43". 
: 0.71684 
: 9.72877 
0.44561 
: 2.7901. 



9. Given a = 34, 
6 = 22, 
B = 30° 20' ; 
find A = 51° 18' 27", 

C = 98° 2r 33", 
c = 43.098, 
^' = 128° 4V 33", 
C" = 20°58 / 27", 
c' = 15.593. 
Here b < a, % but > a sin i?, and 
7? < 90°. 

.-. two solutions. 

loga = 1.53148 
logsin J5 = 9.70332 

colog b = 8.65758 - 10 
log sin ^4 = 9.89238 

A= 51°18 / 27". 
.-.^' = 128° 41' 33". 

,-. C= 98°21 , 33 // . 
t ;G'~ 20° 58' 27", 



98 



PLANE TRIGONOMETRY. 



loga = 1.53148 
log sin = 9.99536 
colog sin A = 0.10762 





logc 


= 1.63446 




c 


= 43.098. 




log a 


= 1.53148 




log sin C 


= 9.55382 




colog sin A 


= 0.10762 




logc' 


= 1.19292 




c' 


= 15.593. 


10 


. Given 




b 


= 19, c - 18 


, C '= 15° 49' ; 


find 


B 


= 16° 43' 13", 




A 


= 147° 27 / 47", 




a 


= 35.519, 




B' 


= 163° 16' 47", 




A' 


= 0° 54' 13", 




a' 


= 1.0415. 


There are two solutions, 


for 


c 


<&, 


but 


c 


> b sin C, 


and 


C < 90°. 




log b 


= 1.27875 




log sin C 


= 9.43546 




colog c 
log sin B 


= 8.74473- 10 




= 9.45894 




B 


= 16° 43' 13". 




r.B' 


= 163° 16' 47". 




..A 


= 147° 27' 47". 




.-.A' 


= 0° 54' 18". 




log b 


= 1.27875 




colog sin B 


= 0.54106 




log sin A 


= 9.73065 



log a = 1.55046 

a = 35.519. 

log 6 = 1.27875 

colog sin & = 0.54106 

log sin J/ = 8. 19784 

log a' = 0.01765 

a' = 1.0415. 



11. Given a = 75, b = 29, B = 
16° 15' 36" ; find the difference be- 
tween the areas of the two corre- 
sponding triangles. 

The triangle which is the differ- 
ence of the two triangles has for its 
altitude a sin jB, and two of its sides 
are of length 29. 

log a = 1.87506 
log sin B = 9.44715 
log (a sin B) = 1.32221 
a sin B = 21. 
29 2 - 21 2 = (29-21)(29 + 21) 

= 8 x 50 = 400. 

... V29 2 - 21 2 = 20. 

Hence, the base of the triangle is 
2 x 20 = 40, and its altitude 21. Its 
area is therefore \ x 40 x 21 = 420. 

12. Given in a parallelogram the 
side a, a diagonal d, and the angle 
A made by the two diagonals ; find 
the other diagonal. 

Special case : a = 35, d = 63, 
A = 21° 36' 30". 

a = 35. 
id = 31.5. 
A = 21° 36' 30". 
colog a = 8.45593- 10 
log id = 1.49831 
log sin A = 9.56615 
log sin B = 9.52039 

B= 19° 2 V 20". 

C = 139° 2' 10". 

loga= 1.54407 

log sin (7 = 9.81663 

colog sin A = 0.43385 

logtd' = 1.79455 

id' = 62.3086. 

d' = 124.617. 



TEACHERS EDITION. 



99 



Exercise XIX. Page 78. 



1. Given find 




log b = 2.94077 


a = 77.99, A = 51° 15', 




log sin A =9.99335 


b = 83.39, B = 56° 30'. 




colog sin B = 0.05924 


C = 72° 15' ; c = 95.24. 




log a = 2.99336 


b + a = 161.38. 




a = 984.83. 


& - a = 5.4. 






J5 + ,4 = 107° 45'. 


3. 


Given find 


*(£ + ^)= 53° 52' 30". 


a - 


= 17, A = 77° 12' 53", 


log (6 -a) =0.73239 
colog (6 + a) = 7.79215-10 


b-- 
C- 


= 12, B = 43° 30' 7", 
= 59° 17' ; c = 14.987. 


lofftan*(£ + ^) = 0.13675 




a + b = 29. 


logtani(B-^l) = 8.66129 




a — 6 = 5. 


$(B-A)= 2° 37' 30". 
A = 51° 15'. 




A + 5 = 120° 43'. 

i(A + B)= 60° 21' 30". 


B = 56° 30'. 




log (a -6)= 0.69897 
colog (a + &) = 8.53760-10 


log 6 = 1.92111 

log sin C = 9.97882 

colog sin 5 = 0.07889 

logc = 1.97882 


log 
log 


tan iU + 5) = 10.24486 
tan* (A -B) = 9.48143 
£(JL - B) = 16° 51' 23". 
A = 77° 12' 53". 


c = 95.24. 




B = 43°30' 7". 
log 6 = 1.07918 


2. Given find 




log sin C = 9.93435 


6 = 872.5, B = 60° 45' 2", 




colog sin B = 0.16217 


c = 632.7, C = 39° 14' 58", 




logc = 1.17570 


^4 = 80°; a = 984.83. 




c = 14.987. 


b - c = 239.8. 






6 + c = 1505.2. 


4. 


Given find 


B + = 100°. 


b 


= V5, B = 93° 28' 36", 


i(£+C) = 50°. 


c 


= V3, C = 50° 38' 24", 


log (6 -c) = 2,37985 


A 


= 35° 53'; a = 1.3131. 


colog (b + c) = 6.82240 -10 




V5 = 2.2361. 


log tan 4 (5 + 0) = 0.07619 




V3 = 1.7321. 


logtani(B- C) = 9.27844 




b + c = 3.9681. 


i(JB- C) = 10°45 / 2". 




&_c = 0.5040. 


£ = 60° 45' 2". 




5 + C = 144° 7'. 


C = 39° 14' 58". 




i(B+ C)= 72° 3' 30". 


L.of; 


, 





100 

log(6-c) = 

colog (b -f c) = 
log tan i (B + C) = 
logtani(J5~ C7)=s 

log c '== 

log sin ^L = 

colog sin C = 

log a — 

a = 



PLANE TRIGONOMETRY. 



9.70243- 10 
9.40142- 10 

10.48973 
9.59358 

21° 25' 6". 

93° 28 / 36". 

50° 38' 24". 

0.23856 
9.76800 
0.11172 
0.11828 

1.3131. 



5. Given 
a = 0.917, 
b = 0.312, 

C = 33° V 9" 



a + b = 

a — b = 

A + B = 

i{A+B) = 

log (a - 6) = 

colog (a + 6) = 

log tan | (A + B) = 

log tan £ (A ~B) = 

i(A-B) = 

A = 

B = 

log 6 = 

log sin C = 

colog sin B = 

lOgC : 



find 
= 132° 18 / 27 
= 14° 34' 24 
= 0.6775. 
1.229. 
0.605. 

146° 52' 51" 

73° 26' 25", 

9.78176 -: 

9.91045 - : 

10.52674 



10.21895 

58° 52' 2". 
132° 18 7 27". 

14° 34' 24". 
9.49415 - 10 
9.73750 
0.59926 



: 9.83091 
: 0.6775. 



10 



6. Given 
a = 13.715, A 
c = 11.214, C 
j? = 15°22'36"; b 

a — c = 



find 
= 118° 55' 49" 
= 45°41 / 35 // 
= 4.1554. 

2.501. 
24,929. 



A + C = 

i(A+C) = 

log(a-c) = 

colog (a + c) = 

log tan i(A + C) = 

log tan i (A - C) = 

i(A-C) = 
A = 

C = 

log a = 

log sin B = 

colog sin A = 

log 6 = 

b = 

7. Given 

b = 3000.9, B 

c = 1587.2, C 

A = 86° 4' 4" ; a 

6 + c = 

6-6 = 

i(£+C) = 

log (6 - c) = 

colog (b + c) = 

logtan£(.B+ 0) = 

logtani(J5- C) = 

i(J5-C) = 

= 

log 6 = 

log sin A = 

colog sin B = 

log a = 



8. Given 
a = 4527, A 

b = 3465, J5 

C ;== 66° 6' 27" ; c 






164° 37' 24 

82° 18' 42' 

0.39811 

8.60330 - 10 
10.86968 

9.87109 

36° 37' 7". 
118° 55' 49". 

45° 4r 35". 
1.13720 
9.42352 
0.05789 
0.61861 
4.1554. 

find 
= 65° 13' 51", 

= 28° 42' 5", 
= 3297.2. 

4588.1. 
1413.7. 
93° 55' 56". 
46° 57' 58". 

3.15036 

6.33837 - 10 
10.02983 

9.51856 
18° 15' 53". 
28° 42' 5". 
65° 13' 51". 
3.47726 
9.99898 
0.04191 
3.51815 
3297.2. 

find 
= 68° 29' 15", 
= 45° 24' 18", 
= 4449, 



TEACHEKS EDITION. 



101 



a - b 

a-b 

A + B 

h(A + B) 

log (a - b) 

colog (a + b) 

log tan* (4 + B) 

log tan i (A - B) 

i(A-B) 

A : 

B 

log a : 
log sin C ■ 
colog sin A 

logC : 



= 7992. 

*= 1062. 

= 113° 53' 33". 

= 56° W 47". 

= 3.02612 

= 6.09734- 10 

= 10.18659 

= 9.31005 

: 11° 32 / 28". 
68° 29' 15". 
45° 24' 18"; 

: 3.65581 

9.96109 
; 0.03136 

3.64826 

4440. 



9. Given 
a — 55.14, 
b = 33.09, 
C = 30° 24' 



find 
A = 117° 24' 32" 

B = 32° 11' 28" 
c = 31.431. 



a + b : 
a — 6 : 



; 88.23. 
22.05. 
149° 36'. 

: 74° 48'. 



log (a - b) 
colog (a + b) 
log tan | ( A + 5) 
log tan $ (A - B) = 9. 96371 



1.34341 

8.05438 - 10 
10.56592 



£: 
lOg 6: 

log sin C : 

colog sin B ■. 

lo^C : 



: 42° 36' 32". 

: 117° 24' 32". 
: 32° 11' 28". 

1.51970 

9.70418 
: 0.27348 
: 1.49736 

; 31.431. 



10. Given 
a = 47.99. 
6 = 33.14, 
C = 175° W 10" 

a + b -. 

a — b -- 

A + B-. 

i(A + B) 

log (a - b) : 

colog (a + b) : 

logtani(^L + £): 

log tan \{A - B) ■. 

\(A-B). 

A: 

B 

log b : 

log sin C ■■ 
colog sin B -. 

lOg C : 



find 
A = 2° 46' 8", 
B= 1°54'42" 
c = 81.066. 

: 81.13. 

: 14.85. 

: 4° 40' 50". 
: 2° 20' 25". 

: 1.17173 
: 8.09082 - 10 
: 8.61138 
: 7.87393 
:<»- 25' 43" 
:2 : 46' 8". 
: 1 - 54' 42". 
: 1.52035 
: 8.91169 
: 1.47680 
: 1.90884 
: 81.066. 



11. If two sides of a triangle are 
each equal to 6. and the included 
angle is 60 3 , find the third side. 

Since a = 6, 

A = B. 
A + B= 120°. 

.-. A = B = C = 60°. 
.-. a =b = c = 6. 

12. If two sides of a triangle are 
each equal to 6. and the included 
angle is 120°, find the third side. 

A + B = 60 3 . 
... A =J? = 30°, 

a = 6 = b. 

log a = 0.77815 

log sin C = 9.93753 

colog sin A = 0.30103 

logc = 1.01671 

c = 10.392. 



102 



PLANE TRIGONOMETRY. 



13. Apply Solution I to the case 
in which a is equal to b ; that is, 
the case in which the triangle is 
isosceles. 

If a — 6, the formula 



tanj(4-i?) = 



x tan $(A + B) 



a-\-b 

becomes 

tan £(.4 - B) = 0. 

.-. A - B = 0. 

A = B 

= i(180°-C) 

= 90° - i C. 

a sin C 

c = 

sin A 

14. If two sides of a triangle are 
10 and 11, and the included angle is 
50°, find the third side. 

a + b = 21. 
a-.— 6 = 1. 
A + B = 130°. 
l(A + B)= 65°. 
log(a-6) = 0.00000 
colog (a + 6) =. 8.67778 -10 
log tan i (A + B) = 10.33183 
log tan £ (4 - JB) = 9.00911 

i(A -B) = 5° 49' 51". 
A = 70° 49 / 51". 
jB = 59°10' 9". 
log 6 = 1.00000 
log sin = 9.88425 
colog sin B = 0.06617 
log c = 0.95042 
c = 8.9212. 

15. If two sides of a triangle are 
43.301 and 25, and the included 
angle is 30°, find the third side. 

a + b = 68.301. 
a - b = 18.301. 



A + B = 150°. 
i(A + B)= 75°. 
log (a -'&) = 1.26247 
colog (a + 6) = 8.16557 - 10 
log tan i(A + B) = 10.57195 
log tan i (A - B) = 9.99999 
i(A - B)= 45°. 
A = 120°. 
B= 30°. 
.-.in isosceles triangle ABC 
c = b = 25. 

16. In order to find the distance 
between two objects A and B sep- 
arated by a swamp, a station C was 
chosen, and the distances CA = 3825 
yards, CB = 3475.6 yards, together 
with the angle A CB = 62° 31', were 
measured. Find the distance from 
A to B. 

b-h a = 7300.6. 
b - ft = 349.4. 
B + A = 117° 29'. 
i(B + A) = 58°44 / 30 // . 
log(6-a) = 2.54332 
colog (b -f a) = 6.13664 - 10 
logtan|(E + -4) = 10.21680 
logtan£(£ - 4) =. 8.89676 

i(B-A)= 4° 30' 29". 
B = 63° 14' 59". 
A = 54° 14' 1". 
log 6 = 3.58263 
log sin = 9.94799 
colog sin B = 0.04916 
log c = 3.57978 

c = 3800. 
.-. 4J5 = 3800 yards. 

17. Two inaccessible objects A 
and B are each viewed from two 



TEACHERS EDITION. 



103 



stations. C and D, on the same side 
of AB and 562 yards apart. The 
angle ACB is 62° 12', BCD 41° 8', 
ABB 60° 49', and ABC 34° 51' ; 
required the distance AB. 




In triangle ACB 

A = 180° - (C + D) 

= 41° 49'. 
6 _ sin 34° 51' 
562 ~ sin 41° 49'* 
a 562 sin 34° 51' 



sin 41° 49' 

log 562 = 2.74974 

log sin 34° 51' = 9.75696 

colog sin 41° 49' = 0.17604 

log b = 2.68274 

b = 481.66. 

In triangle CBB 

B= 180° _(C + D) 

= 43° 12'. 

a _ sin 95° 40' 

562~ sin 43° 12'' 

562 cos 5° 40' 



sin 43° 12' 
log 562 = 2.74974 

log cos 5° 40' = 9.99787 

colog sin 43° 12' = 0.16460 

log a = 2.91221 

a = 816.98. 

In triangle ACB 

Uuii(A-B) = C —xtMii(A + B) 
a + b 



i(A + B). 

a — b ■- 
a + b: 

log (a - b) : 

colog (a + b) ■- 

log tan i {A + B) ■. 

log tan i (A - B) 



: } (180° - C) 
: 58° 54'. 

: 816.98-481.66 
: 335.32. 

: 816.98 + 481.66 
: 1298.64. 



2.52548 

6.88651 

10.21951 



10 



i(A 



A 



= 9.63150 

23° 10' 26" 
82° 4' 26" 



log a = 2.91221 

log sin C = 9.94674 

colog sin A — 0.00417 

logc = 2.86312 

c = 729.67. 

.-. AB = 729.67 yards. 

18. Two trains start at the same 
time from the same station, and 
move along straight tracks that 
form an angle of 30°, one train at 
the rate of 30 miles an hour, the 
other at the rate of 40 miles an 
hour. How far apart are the trains 
at the end of half an hour ? 

a + b = 35. 
a — b = 5. 
A + B = 150°. 
l(A + B)= 75°. 

log (a -b) = 0.69897 
colog (a + b) = 8.45593 - 10 
\ogtzni(A+B) = 10.57195 
log tan i (A - B) = 9. 72685 

i (A - B) = 28° 3' 52". 
B= 46° 56' 8". 
A = 103° 3' 52". 



104 



PLANE TRIGONOMETRY. 



log 6 = 1.17609 

log sin C= 9.69897 

colog sin B = 0.13633 

logc = 1.01139 

c = 10.266. 

Therefore, the trains are 10.266 
miles apart. 

19. In a parallelogram given the 
two diagonals 5 and 6, and the 
angle that they form 49° 18'; rind 
the sides. 

In the parallelogram ABBE 
let EB = 6, and AD = 5, 

and Z BCA = 49° 18'. 



In triangle ACB 



let 



2.5. 



Find 



BC = a. 

AC = b: 
AB = c. 
a-b = 0.5. 
a + b = 5.5. 
A + B = 130° 42'. 
\(A + B)= 65° 21'. 
log (a -&)= 9.69897-10 
colog (a + 6) = 9.25964 - 10 
log tan £ (A + B) = 10.33829 
log tan i (4 - JS) = 9.29690 

i(4 - J5) = 11° 12' 20". 
^L = 76° 33' 20". 
J5 = 54° 8' 40". 
log a = 0.47712 
log sin = 9.87975 
colog sin A = 0.01207 
logc = 0.36894 

c = AB = 2.3385. 
In triangle AEC 

EC = a = 3, 
AC = b = 2.5, 
Z ACE = 130° 42'. 



A + J?: 

i(^ .+ ■») = 

log (a - b) : 

colog (a + 5) : 

logtan£(^l + J£) • 

log tan i (A — E) ■ 

i(A-E): 
A I 

log a: 

log sin C ■■ 

colog sin A : 

lOgC: 
C = EA : 



: 49° 18'. 
24° 39*. 
9.69897 - 10 
9.25964 - 10 

: 9.66171 

: 8.62032 

2°23 / 20 // . 
:27° 2' 20". 
: 0.47712 
: 9.87975 
: 0.34238 
: 0.69925 
: 5.0032. 



20. In a triangle one angle is 
139° 54', and the sides forming the 
angle have the ratio 5 : 9. Find the 
other two angles. 

a = 9. 
6 = 5. 
a + b = 14. 
a - b = 4. 
JL+ J5 = 40°6'. 
£ (A + 5) = 20° 3'. 

log (a - 6) = 0.60206 
colog (a + 6) = 8.85387 - 10 
logtan£(^. + B) = 9.56224 
log tan -J (4 - #) = 9.01817 

i (A - B) = 5° 57' 10". 
A = 26° 0' 10". 
£ = 14° 5' 50". 

21. In order to find the distance 
between two objects A and B 
separated by a pond, a station C 
was chosen, and the distances CA 
= 426 yards, CB = 322.4 yards, 
together with the angle ACB = 
68° 42', were measured. Find the 
distance from A to B. 



TKACHEKS EDITION. 



105 



b + a = 748.4. 

b - a = 103.6. 

# + .4 = 111° 18'. 

t(B + A)= 55° 39'. 

log (b - a) = 2.01530 
colog (b + a) = 7.12587 - 10 
logtan£(£ + ^4) = 10.16530 
logtan£(7^-^4) = 9.30653 



i(B~ A) = 11° 27' 1". 

.-. 7? = 07° 6' 1". 

log 6 = 2.62941 

log sin C = 9.96927 

colog sin i? = 0.03565 

log c = 2.63433 

c =430.85. 
.-. .422 = 430.85 yards. 



Exerctse XX. Page 83. 



1. Given a = 51 , b = 65, c = 20 
find the angles. 

a= 51 

6= 65 

c = 20 

2 s = 136 

8 = 68. 

s — a = 17. 

s - 6 = 3. 

s _ c = 48. 

colog s = 8.16749 - 10 
colog (s - a) = 8.76955 - 10 
log(s-6) = 0.47712 
log (s - c) = 1.68124 

2 )19.09540-20 
log tan i A = 9.54770 

iA = 19° 26' 24". 
A = 38° 52' 48". 

colog s = 8.16749-10 
colog (s -b)= 9.52288 - 10 
log's- a) = 1.23045 
log (« - c) = 1.68124 

2 )20.60206 -2 
log tan i 5 = 10.30103 

\B = 63°26 / 6". 
B = 126° 52' 12". 
^1 + B = 165° 45'. 
.-. C = 14° 15'. 



2. Given a = 78. 6 = 101, c = 29; 
find the angles. 

a = 78 

6 = 101 

c= 29 
2 s = 208 

8 = 104. 
s-a= 26. 
s — b = 3. 
s — c — 75. 

colog s = 7.98297 - 10 
colog (s - a) = 8.58503 - 10 
log (8 -o)= 0.47712 
log (s - c) = 1.87506 

2 )18.92018-20 
log tan i A = 9.46009 

±A = 16° 5 / 27". 
A = 32° 10' 55". 

colog s= 7.98297-10 
colog (s- 6)= 9.52288- 10 
log (8- a) = 1.41497 
log (s - c) = 1.87506 

2 )20.79588-20 
log tan £5 = 10.39794 

* B = 68° ll 7 55". 

B = 136° 23' 50". 

A + B = 168° 34' 45". 

.-. C = 11° 25' 15". 



106 



PLANE TRIGONOMETRY. 



3. Given a = 111, 6=145, c=40 ; 
find the angles. 

a = 111 
6=145 
c= 40 
2 s = 296 
s = 148. 
s - a = 37. 
s - 6 = 3. 
s - c = 108. 
colog s = 7.82974-10 
colog (s - a) = 8.43180 - 10 
log(s-6) = 0.47712 
log (s - e) = 2.03342 
_J 2 )18.77208-20 

log tan £^1 = 9.38604 

iA = 13° 40' 16". 
A = 27° 20' 32". 

colog s = 7.82974-10 
log(s-a) = 1.56820 
colog (s- 6) = 9.52288-10 
log (s - c) = 2.03342 

2 )20.95424-20 
log tan iB = 10.47712 

$B= 71°33 / 54 // . 

B = 143° V 48". 

5 + A = 170° 28' 20". 

.-. G = 9° 3r 40". 

4. Given a = 21, 6 = 26, c = 31 ; 
find the angles. 

a = 21 

6 = 26 

c = 31 
2s = 78 

s = 39. 
s - a = 18. 
s - 6 = 13. 
s - c = 8. 



colog s = 8.40894-10 
colog(s-a) = 8.74473-10 
log (s -6)= 1.11394 
log (s - c) = 0.90309 

2 )19.17070-20 
log tan i A = 9.58535 

\A =21°3 / 6.3". 
.-. .A = 42° 6' 13". 

colog s = 8.40894-10 
log(s-a) = 1.25527 - 
colog (s- 6)= 8.88606-10 
log (s - c) = 0.90309 

2 )19.45336 -20 
logtani£= 9.72668 

1^ = 28° 3' 18". 

.-.5 = 56° 6' 36". 

A + £ = 98° 12' 49". 

.-. O = 81° 47 / 11". 



5. Given a = 19, 6 = 34, c = 49 ; 
find the angles. 

a= 19 

6= 34 

c= 49 

2s = 102 

s = 51. 
s - a = 32. 
s - 6 = 17. 
s - c = 2. 

colog s= 8.29243- 10 
colog (s - a) = 8.49485 - 10 
log(s-6)= 1.23045 
log (s - c) = 0.30103 

2 )18.31876-20 
logtan£X= 9.15938 

iA= 8°12 / 48". 
A = 16° 25' 36". 



TEACHERS' ED IT TON. 



107 



colog .s = 8.20243 - 10 
\og(s — a) = 1.50515 
colog («-&) = 8.76955-10 
log (8 - c) = 0.30103 

2 )18.86816 - 20 
logtan|£ = 9.43408 

±B = 15 6 12'. 
B= 30°24 / . 
.-. G = 133° 10' 24". 



6. Given a = 43, b = 50, c : 


= 5 


find the angles. 




a= 43 




6= 50 




c — 57 
2s = 150 




s = 75. 




s - a = 32. 




s - b = 25. 




s - c = 18. 




colog s=8. 12494 - 


10 


colog (s- a) = 8.49485- 


10 


log («-&) = 1.39794 




\og(s — c) = 1.25527 




2)19.27300- 


20 


logtani^4 = 9.63650 




|.4 = 23 24 / 47.6". 


^ = 46° 49' 35" 




colog s= 8.12494- 


10 


log(s-a) = 1.50515 




colog (s - b)= 8.60206- 


10 


log(s — c) = 1.25527 




2)19.48742 - 


20 


logtani^= 9.74371 




£E = 28°59 / 52" 




B = 57° 59' 44" 




p .-. C = 75° 10' 41" 




7. Given a = 37, 6 = 58, c : 


= 7 


find the angles. 





79; 



a= 37 

b= 58 

c = 79 

2s = 174 

s = 87. 
s — a = 50. 
s - b = 29. 
8 - c = 8. 

colog 8= 8.06048 - 10 
colog(s-a) = 8.30103-10 
log(«-o) = 1.46240 
log (s - c) = 0.90309 

2 )18.72700 -20 
logtani^l= 9.36350 

£4 = 13° 0' 14.5". 

A = 26° 0' 29". 

colog s = 8.06048 - 10 
log(s-a) = 1.69897 
colog (s- b)= 8.53760-10 
log (a - c) = 0.90309 

2 )19.20014-20 
logtan|£ = 9.60007 

\B = 21°42 / 40". 

B= 43° 25' 20". 

.-. C = 110° 34' 11". 

8. Given a = 73, b = 82, c = 91 
find the angles. 

a= 73 
6= 82 
c= 91 

2 s = 246 

s = 123. 
s — a = 50. 
s -6 = 41. 
s - e = 32. 



108 



PLANE TRIGONOMETRY. 



colog s = 7.91009-10 
co\og(s-a)= 8.30103-10 
log(s-6) = 1.61278 
log (s - c) = 1.50515 

2 )19.32905-20 
log tan i ^4 = 9.66453 

i^. = 24°47 / 29 // . 
A = 49° 34' 58". 

colog s = 7.91009-10 
log (s- a) = 1.69897 
colog (s-&) = 8.38722-10 
log (s - c) = 1.50515 

2 )19.50143-20 
logtan£jB = 9.75072 

15 = 29° 23' 29". 

B = 58° 46' 58". 

.-. C = 71° 38' 4". 



9. Given a = 14.493, 6 = 55.4363, 
= 66.9129 ; find the angles. 

a = 14.493 
6= 55.4363 
c= 66.9129 



2 s = 136.8422 

s = 68.4211. 
s - a = 53.9281. 
s~b= 12.9848. 
s - c = 1.5082. 

colog s = 8.16481-10 
colog (s- a) = 8.26819-10 
log(s-6) = 1.11343 

lOg (S - C) = Q.17846 

2 )17.72489-20 
logtani^i = 8.86245 

i^i = 4°10'0.7". 
^1 = 8° 20' 1". 



colog s = 8.16481-10 
log(s-a)= 1.73181 
colog (s-&) = 8.88657 -10 
log (s - c) = 0.17846 

2 )18.96165-20 
logtan££ = 9.48082 

iJ5= 16° 50' 2.6". 

B= 33°40 / 5". 
.-. C = 137° 59' 54". 

10. Given a= Vs, 6= Ve, c= V7; 
find the angles. 

a =V5 = 2.2361 

6 = V6 = 2.4495 

c = V 7 = 2.6458 

2s = 7.3314 

s = 3.6657. 
s - a = 1.4296. 
s - b = 1.2162. 
s - c = 1.0199. 

colog s = 9.43585-10 
colog (s-a)= 9.84478 - 10 
log («-&)= 0.08500 
log (s - c) = 0.00856 

2 )19.37419-20 
log tan i A = 9.68709 

£J. = 25°56 / 36". 
A = 51° 53' 12". 

colog (s-&) = 9.91500-10 
log(s-c)= 0.00856 

colog s = 9.43585-10 
log (s - a) = 0.15522 

2 )19.51463-20 
log tan £.8= 9.75732 

i# = 29°45 / 54". 
£ = 59° 3r 48". 
.-. C = 68° 35'. 



TEACHERS' EDITION. 



109 



11. Given a = 6, 6 = 8, c = 10 ; 
find the angles. 

a= 6 
b= 8 
c = 10 
2s = 24 
s= 12. 
8 — a = 6. 
s _ 6 = 4. 
s - c = 2. 
cologs = 8.92082 - 10 
colog(s-a) = 9.22185-10 
log(s-6) = 0.60206 
log(s-c) = 0.30103 

2)19.04576 - 20 
logtani^l = 9.52288 

iA = 18° 26' 6". 
A = 36° 52' 12". 
Since 10 2 = 6 2 + 8 2 , the triangle 
is a right triangle, 
and C = 90°. 

.-. B = 53° V 48". 

12. Given a = 6, 6 = 6, c = 10 ; 
find the angles. 



a= 6 




b= 6 




c = 10 




2 s = 22 




s = 11. 




s — a = 5. 




s - 6 = 5. 




s — c = 1. 




cologs = 8.95861- 


-10 


log(s-a) = 0.69897 




log (8 -6)= 0.69897 




colog (s-c)= 0.00000 




2)20.35655- 


-20 


logtaniC= 10.17827 





iC= 56° 26' 33". 
C = 112° 53' 6". 
Since this is an isosceles triangle, 
A = B = 1(180° - C) 
= 33° 33' 27". 

13. Given a = 6, 6 = 6, c = 6 ; 
find the angles. 

The triangle is equilateral and is 
therefore also equiangular. 

.:A = B=C = ioi 180° = 60°. 



14. Given a = 
find the angles. 

a •■ 

6: 

C : 

2S: 

S : 

s — a ■ 

8-6: 

S — C ■ 
COlOg 8 : 

colog s — a ■■ 

log S — 6 : 

log S — C 
c 

log tan i A ■■ 
IA 

A: 

COlOg S : 

log S — a : 

COlOg S — b : 
lOg S — C : 

2 
log tan i B : 

.'. C: 



6, 6 = 9, c = 12 ; 

= 6 
= 9 
= 12 

= 27 

= 13.5. 

= 7.5. 
= 4.5. 
= 1.5. 

= 8.86967 - 10 
= 9.12494-10 
= 0.65321 
= 0.17609 
2 )18.82391 -20 
9.41196 - 10 
: 14° 28' 39". 
28° 57' 18". 
8.86967 - 10 
0.87506 
9.34679 - 10 
0.17609 
2 )19.26761-20 
: 9.63380-10 

23° 17' 3". 

46° 34' 6". 

104° 28' 36". 



110 



PLANE TRIGONOMETRY. 



15. Given a = 2, b = Ve, c 
V3 — 1 ; find the angles. 

a = 2. 

6 = Vis = 2.44947 

c = V3 - 1 = 0.73205 
2s = 5.18152 

s = 2. 59076. 
s-a = 0.59076. 
s- 6 = 0.14129. 
s-c = 1.85871. 



log(s-a) 

log (s - 6) 

log (s - c) 

colog s 

logr 2 



9.77141 - 10 
9.15011 - 10 
0.26921 
9.58657 - 10 
18.77730-20 

logr = 9.38865- 10. 



log tan $ A -. 
log tan •} B • 
log tan i C = 

\A: 

iB: 

iC: 

A: 

B: 

C: 



9.61724. 

10.23854. 

9.11944. 

22° 30'. 
60°. 
7° 30'. 
45°. 
120°. 
15°. 



16. Given a = 2, b = Ve, 
V3 + 1 ; find the angles. 

a = 2. 
b = V6 = 2.44947 
c=V3 + 1 = 2.73205 
2s = 7.18152 

s = 3.59076. 
* -a = 1.59076. 

s-6 = 1.14129. 
s _ c = 0.85871. 



c = 



log (« - a) = 0.20160 

log (s - 6)= 6.05740 

log(s-c)= 9.93385-10 

colog s = 9.44481 - 10 

logr 2 = 19.63766-20 

logr= 9.81883 - 1( 

log tan iA = 9.61723. 
logtani^ = 9.76143. 
log tan* C = 9.88498. 

-M = 22°30 / . 
iB = S0°. 
i C = 37° 30'. 

A = 45°. 
£ = 60°. 

C = 75°. 



17. The distances between three 
cities A, B, and C are as follows : 
AB=165 miles, ^.0=72 miles, and 
i?C= 185 miles. B is due east from 
A. In what direction is C from J.? 
What two answers are admissible? 





a = 


185 






b = 


72 






c = 


165 






2s = 


422 






s = 


211. 




s 


— a = 


26. 




s 


-6 = 


139. 




s 


-c = 


46. 




colog s = 


7.67572 - 


10 


colog (s - 


-a) = 


8.58503 - 


10 


log(s 


-&) = 


2.14301 




log(s 


-€) = 

2) 


1.66276 






20.06652 - 


20 


log tan 


10.03326 






M = 


47° 11' 31" 






^4 = 


94° 23' 2" 





TEACHERS' EDITION. 



Ill 



Angle BAC = 94° 23' 2". Sub- 
tract 90° of the quadrant E to N, 
and we obtain 4° 23' 2" W. of N. 

But C may be to the southward 
of A. Hence two answers are ad- 
missible : W. of N. or W. of S. 

18. Under what visual angle is 
an object 7 feet long seen by an 
observer whose eye is 5 feet from 
one end of the object and 8 feet 
from the other end ? 

a — 5 
b= 8 

c = J_ 
2s = 20 
s = 10. 
s — a = 5. 
8-6 = 2. 
s - c = 3. 
colog s = 9.00000-10 
log (8- a) = 0.69897 
log (8 -6)= 0.30103 
colog (s - c) = 9.52288 - 10 
2 )19.52288 -20 
logtanJC= 9.76144 

I C = 30°. 

C = 60°. 

19. When Formula [28] is used 
for rinding the value of an angle, 
why does the ambiguity that occurs 
in Case II not exist ? 

When Formula [28] is used for 
finding the value of an angle, the 
ambiguity that occurs in Case II 
does not exist because the sides are 
all known and the angle can have 
but one value ; while in Case II the 
side opposite the angle is not known, 



and may have two values, and there- 
fore the angle also may have two 
values. 



20. If the sides of a triangle are 
3, 4, and 6, find the sine of the 
largest angle. 





a = 


3 




b = 


4 




c = 
2s = 


6 
13 




s — 


6.5. 


s 


— a = 


3.5. 


s 


-b = 


2.5. 


8(8 - 


-c) = 


3.25. 



log (s- a) = 0.54407 
log (s -6)= 0.39794 
colog s(s - c) = 9.48812 - 10 
2 )20.43013-20 
logtaniC= 10.21507 

iC= 58°38 , 25 ,/ . 
C = 117° W 50". 
log sin (7 = 9.94879. 
sin C = 0.88877. 



21. Of three towns A, J5, and C, 
A is 200 miles from B and 184 miles 
from C, B is 150 miles due north 
from C. How far is A north of C ? 

a = 150 

b= 184 

c = 200 

2 s = 534 

s = 267. 
s — a = 117. 
s - b = 83. 
s - c = 67. 



112 



PLANE TRIGONOMETRY. 



colog s = 7.57349-10 
log (8- a) = 2.06819 
log (*-&)= 1.91908 
colog (s - c) = 8.17393-10 
2 )19.73469-20 
logtaniC= 9.86735 

1 C = 36° 22' 58". 
C = 72° 45' 56". 

Draw ± from A to £C. To find 
a'(part cut off by _1_ on BC from J.). 

a' = b cos C. 

log 6 = 2.26482 

log cos C = 9.47171 

loga' = 1.73653 

a' = 54.516. 

Therefore, A is 54.516 miles north 
of C. 



22. The sides of a triangle are 
78.9, 65.4, 97.3, respectively. Find 
the largest angle. 

a= 78.9 

b = 65.4 

c = 97.3 

2 s = 241.6 

s = 120.8. 
s - a = 41.9. 
s-b = 55.4. 
s - c = 23.5. 

colog s = 7.91793-10 

log (8- a) = 1.62221 

log (8 -6)= 1.74351 

colog (s - c) = 8.62893 - 10 

2 )19.91258-20 

logtan£C = 9.95629-10 

i C = 42° V 17". 
C = 84° 14' 34". 



23. The sides of a triangle are 
487.25, 512.33, 544.37, respectively. 
Find the smallest angle. 

a = 487.25 

b= 512.33 

c= 544.37 

2s = 1543.95 

s = 771.975. 
s -a = 284.725. 
s - 6 = 259.645. 
s-c = 227.605. 

colog s= 7.11239-10 
colog (s- a) = 7.54557-10 
log (8 -6)= 2.41438 
log (s - c) = 2.35718 

2 )19.42952-20 
log tan £.4=: 9.71476-10 

iA = 27° 24' 27". 
^4 = 54° 48' 54". 

24. Find the angles of a triangle 
V3 + 1 V3-1 



whose sides are 
2 



2 V2 2 V2 



respectively. 

V3 + 1 V6+V2 



2V2 



2.44947 + 1.41421 



3.86368 



: 0.96592. 



V3_ 1 _ V6 -V2 

2V2 4 



2.44947 - 1.41421 



1.03526 



= 0.25882. 



V3 1.73205 



: 0.86603. 



TEACHERS EDITION. 



113 



a 

b 
c 
2s = 2.090 



0.96592 

0.25882 
0.86608 



s ■ 
s — a 

S-b: 

S — C : 

log(s- a) : 
log(s - b) 

log (8 - C): 
COlog S : 

log r 2 : 
log r : 

log tan -J- A ■■ 
log tan i B : 
log tan -J- C ■ 

$A: 
iB: 

iC 

A : 

B 

C 



1.04538. 

: 0.07946. 

0.78656. 

0.17935. 

8.90015 - 10 

; 9.89573 - 10 

9.25370 - 10 

9.98072 - 10 



18.03030 
9.01515 - 



-20 
10. 



10.11500- 10. 
9.11940- 10. 
9.76145 - 10. 

52° 30 7 . 
7° 30 r . 
30°. 

105°. 
15°. 
60°. 



25. The sides of a triangle are 
14.6 inches, 16.7 inches, and 18.8 
inches, respectively. Find the 
length of the perpendicular from 



the vertex of the largest angle upon 
the opposite side. 

a = 18.8 
6 = 16.7 
c = 1^6 
2s = 50.1 
s = 25.05. 
s -a — 6.25. 
s -b = 8.35. 
s-c= 10.45. 
cologs = 8.60119 - 10 
log(s-a)= 0.79588 
colog(s -b)= 9.07831 - 10 
log (s - c) = 1.01912 

2 )19.49450 -20 
log tan ^B = 9. 74725 - 10 
iB = 29° H'46". 
B = 58° 23 / 32". 
Let AD be the perpendicular from 
A upon BC. 



AD 



= sin B. 



14.6 

.-. AD= 14.6 sin B. 
log 14.6= 1.16435 
log sin B = 9.93026 
log AD = 1.09461 
AD = 12.434. 
Therefore, the required perpen- 
dicular is 12.434 inches. 



Exercise XXI. Page 87. 



1. Given a = 4474.5, b = 2164.5. 
C = 116° 30 7 20"; find the area. 
F = iab sin C 
log a = 3.65075 
log b = 3.33536 
colog2 = 9.69897 - 10 
log sin C = 9.95177 
log F= 6.63685 
F = 4,333,600. 



2. Given b = 21.66. c = 36.94, 
A = 66° 4 r 19" ; find the area. 
F = ibcsmA. 
log 6= 1.33566 
logc = 1.56750 
colog2 = 9.69897 - 10 
logsin^L = 9.96097 
log F= 2.56310 
F= 365.68. 



114 



PLANE TRIGONOMETRY. 



3. 


Given a = 510, c = 173, 


B = 


logs = 1.65321 


162° 


30' 28" ; find the area. 




log(s - a) = 0.69897 




loga = 2.70757 
logc =2.23805 
log sin B= 9.47795 
colog2 = 9.69897 - 


10 


log(s-&) = 1.50515 

log(s-c) =0.90309 

2)4.76042 

log F= 2.38021 




log F = 4.12254 




F = 240. 




F = 13,260. 




6. Given a = 624, b = 205, c = 445 
find the area. 

a= 624 


4. 


Given a = 408, & =41, c = 


= 401; 


6= 205 


find 


the area. 




c= 445 




a = 408 




2 s = 1274 




6 = ^41 




s = 637. 




c = 401 


~*- j 


8 - a = 13. 




2s = 850 




s - 6 = 432. 




s = 425. 
s - a = 17. 

s - b = 384. 
s - c = 24. 

logs = 2.62839 
log(s- a) = 1.23045 
log (*,-&) =2.58433 
log (s-c) =1.38021 




s - c = 192. 

logs = 2.80414 

log(s-a) = 1.11394 

log(s-o) =2.63548 

log (s-c) =2.28330 

2 log 1^=8.83686 

log F= 4.41843 

F= 26,208. 




2)7.82338 


*_ 


- 7. Given 6 = 149, ^L = 70° 42' 30' 




^2^=3.91169 




B = 39° 18' 28"; find the area. 




F = 8160. 




A = 70° 42' 30". 

B = 39° 18' 28". 

.-. O = 69° 59' 2". 


5. Given a = 40, b = 13, c : 
find the area. 


= 37; 


log 6 = 2.17319 
log sin A = 9.97490 




a = 40 




colog sin B = 0.19827 




b = 13 




log a = 2.34636 




c =37 




colog 2 = 9.69897- 10 




2 s = 90 




loga = 2.34636 




s = 45. 




log b =2.17319 




8 — a — 5. 




log sin = 9.97294 




s - b = 32. 




log F= 4. 19146 




s - c = 8. 




F= 15,540. 



TEACHERS EDITION. 



115 



8. Given a = 215.0, c = .307.7, 
A = 25° 9' 31" ; find the area. , 

a<c and >csinyl. 
^4 < 90°. .*. two solutions, 
log c = 2.48813 
log sin ^L = 9.62852 

colog a = 7.66575 - 10 
log sin C= 9.78240 

C= 37°17 / 38 // . 
,\ £ = 117° 32' 51". 
Or, C = 142° 42' 22". . 

.-. B' = 12° 8' 7". 
colog 2 = 9.69897 - 10 
log a = 2.33425 
logc =2.48813 
log sin B = 9.94774 
log F = 4.46909 
F = 29,450. 
colog 2 = 9.69897 - 10 
log a = 2.33425 
logc =2.48813 
log sin W = 9.32268 
log F' = 3.84403 

f = 6982.8. 

9. Given b = 8, c = 5, A = 60° ; 
find the area. 

F = ibcs'mA 

= 1(8 x 5) (0.86603) 
= 20 x 0.86603 
= 17.3206. 

10. Given a = 7, c = 3, A = 60° ; 
find the area. 

logc = 0.47712 
log sin J. = 9.93753 

colog a = 9.15490 - 10 
log sin C = 9.56955 

C = 21° 47' 12". 
.-. B = 98° 12' 48". 



colog 2 = 9.69897 

log a = 0.84510 

logc = 0.47712 

log sin B = 9.99552 

log.F: 



10 



1.01671 
F = 10.392. 

11. Given a = 60, B =40° 35' 12", 
area = 12 ; find the radius of the 
inscribed circle. 

\ ac sin B = 12. 
24 



a sin B 
log 24 = 1.38021 
colog a = 8.22185 - 10 
colog sin B = 0. 18669 

logc = 9.78875 - 10 

c= 0.61483. 
a- c = 59.38517. 
a + c = 60.61483. 
A + C = 139° 24' 48". 
J (A + C) = 69° 42 / 24". 

log(a-c) = 1.77368 
colog(a + c)= 8.21742-10 
log tan £ (4 +~ C) = 10.43206 
log tan i (4 - C) = 10.42316 

i(4 - C)= 69° 19' 19". 
.-. A = 139° V 43". 

a sin i> 
b = - 



sino- 
log a = 1.77815 
log sin 5 = 9.81331 
colog sin A = 0.18331 

log B = 1.77477 
b = 59.534. 

a= 60. 
6 = 59.534 
c= 0.61483 
2 s = 120.14883 



116 



•LANE TRIGONOMETRY. 



.-. r 



s = 60.07442. 
F= rs. 

F 12 

s 



60.07442 
log.F = 1.07918 
colog s = 8.22131 - 10 
logr = 9.30049 -10 

r = 0.19975. 

12. Obtain a formula for the area 
of a parallelogram in terms of two 
adjacent sides and the included 
angle. 

By Geometry, area of parallel- 
ogram = base x height. 
In this case, area = bh. 
But h = a sin A. 
.-. area of O = ab sin A. 

13. Obtain a formula for the 
area of an isosceles trapezoid in 
terms of the two parallel sides and 
an acute angle. 

Let a — greater base, 

6 = smaller base, 
h = altitude, 
p = i(a -6), 
and A = angle at lower 

base. 
F=i(a + b)k. 



Now 
But 



h 



— tan A. 



i(a-b) 

.-. h = i(a — b) tan A. 
.-. F = i(a + b) x i (a - b) tan A 
= i(a 2 - b 2 )tsa\A. 

14. Two sides and included angle 
of a triangle are 2416, 1712, and 
30°; and two sides and included 
angle of another triangle are 1948, 
2848, and 150°; find the sum of 
their areas. 



Let a ■ 



: 2416, c 
F-. 

log a : 

lOgC : 
COlOg 2 : 

log sin B -. 

\0gF: 

F- 



Let a 7 = 1948, c': 
F' 

log of ■ 

lOg C' : 
COlOg 2 : 

log sin B' ■ 

lOg J": 
F / 

.-. F+ F' 



= 1712, B = 30° 
: | ac sin B. 

: 3.38310 
: 3.23350 
: 9.69897 - 10 
: 9.69897 
: 6.01454 
= 1,034,000. 

:2848, JB^ISQP 
= i a'c' sin B'. 

: 3.28959 
: 3.45454 
: 9.69897 - 10 
: 9.69897 
: 6.14207 

= 1,387,000. 
: 2,421,000. 



15. The base of an isosceles tri- 
angle is 20, and its area is 100 -r- v3 ; 
find its angles. 

a = b. 

c = 20. 

F = 100 -*- Vs. 
100 

Vs' 

100 

V3' 

10 

V3' 

tan^l. 



\ch~- 



10A: 



h=- 



h_ 

ic'' 



log h = 0.76144 
colog ic = 9.00000 - 10 
log tan A = 9.76144 

A = 30°. 
.-. B = 30°. 
.-. C = 120°. 



teachers' editiox. 



117 



16. Show that the area of a quad- 
rilateral is equal to one-half the 
product of its diagonals into the 
sine of their included angle. 

Let the lengths of the diagonals 
be denoted by a and 6, and the 
included angle by C. Let the lengths 
of the segments made by their point 
of intersection at C be denoted by 
cii, a 2 , and &i, b 2 , respectively. 



Now the area of the quadrilateral 
is equal to the sum of the areas of 
the four triangles. 

.-. F = \ a x b 2 sin C + £ a 2 b 2 sin C 

-f i a 2 bi sin C + \ a{b\ sin C 
— ±(ciib 2 + a 2 b 2 + a 2 bi -f ai&i)sin C 
=i(ai + a 2 )(b 1 + b 2 )sm C 
= iab sin C. 



Exercise XXII. Page 88. 



1. From a ship sailing down the 
English Channel the Eddystone was 
observed to bear N. 33° 45 y W., 
and after the ship had sailed 18 
miles S. 67° 30 / W. it bore N. 11° 
15' E. Find its distance from each 
position of the ship. 




Let A represent the Eddystone, 
C the first position of the ship, and 
B the second. 

a = 18 miles. 
ACE = ^°W. 
DCB = 67° 3CT. 
ABF= 11° 15'. 



ACB = 180° -(ACE + DCB) 

= 78° 45'. 
CBD = 90° - DCB = 22° 30. 
ABC = 90° - (CBD + ABF) 
= 56° 15'. 
.. BAC = 45°. 

b _ sin B 
a sin A 

log a — 1.25527 

log sin B = 9.91985 

colog sin A — 0. 15051 

log b = 1.32563 

6 = 21.166. 

c _ sin C 

a sin A 

\oga= 1.25527 

log sin C = 9.99157 

colog sin A = 0.15051 

logc = 1.39735 

c = 24.966. 

Therefore, the required distances 
are 21.166 miles and 24.966 miles. 

2. Two objects A and B were ob- 
served from a ship to be at the same 
instant in a line bearing N. 15° E. 
The ship then sailed northwest 



118 



PLANE TRIGONOMETRY. 



5 miles, when it was found that 
A bore due east and B bore north- 
east. Find the distance from A 
to B. 

N 




'/ 45° 


105° 


\45° 


75° 


\ 




\ 


I 


\ 


i 


V 


1 

i 

1 




\60 o/ 



Let A and B represent the objects, 
S and S' the first and second posi- 
tions of the ship. 

S' A _ am ASS' 
SS' ~ sin S'AS' 
log SS' = 0.69897 
log Bin ASS' = 9.93753 
colog sin SAS' = 0,01506 
log'S' A = 0.65156 
AB _ sin BS'A 
WA ~ sin S'BA ' 
logS'A = 0.65156 
log sin BS' A = 9.84949 
colog sin S'BA = 0.30103 
log AB = 0.80208 
AB = 6.3399. 
Therefore, the distance from A to 
B is 6.3399 miles. 



3. A castle and a monument 
stand on the same horizontal plane. 
The angles of depression of the top 
and the bottom of the monument 
viewed from the top of the castle 
are 40° and 80° ; the height of the 
castle is 140 feet. Find the height 
of the monument. 




EC = height of castle. 
AB = height of monument. 

MOB = 40°. 
HCA = 10°. 
H AC = 80°. 
HC = 140 feet. 

sin A 
log 140 = 2.14613 
colog sin A = 0.00665 
log^LC = 2.15278 
HCA = 10°, 
MCB = 40°. 
.-. ACB = 40°, 
CAB = 10°. 
.-. ABC = 130°. 

^ICsinC 



AB. 



sin B 



log -4(7 = 2.15278 

log sin = 9.80807 

colog sin B = 0.11575 

log AB= 2.07660 

AB= 119.29. 



TEACHERS EDITION 



119 



Therefore, the height of the 
nonument is 119.29 feet, 

4. If the sun's altitude is 60°, 
what angle must a stick make with 
he horizon in order that its shadow 

m a horizontal plane may be the 
longest possible ? 

The shadow of the stick will be 
the longest when the stick is per- 
pendicular to the rays of the sun. 

Let BC represent the stick, and 
A C the horizontal plane. 
B = 90°. 
A = 60°. 
.-. C = 30°. 

5. If the sun's altitude is 30°, find 
the length of the longest shadow cast 
on a horizontal plane by a stick 10 
feet in length. 

Let a be a stick J_ to rays of sun, 
and c be the longest shadow. 

- = sin A = l. 
c 

c = 2 a = 20. 

Therefore, the longest shadow is 

20 feet. 

6. In a circle with the radius 3 
find the area of the part comprised 
between parallel chords whose 
lengths are 4 and 5. (Two solu- 
tions. ) 

In triangle BOC, 

h = V32 - 22 

= V5. 
F=lx V5 x 4 
= 2 V56 
= 4.4722. 




sin i BOC = f. 

log 2 = 0.30103 
colog 3 = 9.52288 - 10 
log sin ii?OC = 9.82391 

i BOC = 41° 48 / 38". 

BOC = 83° 37 / 16". 

area O = it x 3' 2 . 

logTr = 0.49715 

log 3 2 = 0.95424 

log area O = 1.45139 

area© = 28.274. 
Area of sector BOC 
83° 37' 16" 





360° 


x 28.27 


4 




_ 301036 
" 1296000 


x 28.274 






= 7525 ° x 28.274. 
324000 






log 75259 


= 4.87656 






log 28.274 


= 1.45139 






colog 324000 


= 4.48945 


- 10 




log area 


= 0.81740 




Ai 


ea of sector i>OC 








= 6.5675. 




Ai 


ea of segment ByC 








= 6.5675- 


4.4722 






= 2.0953. 





120 



PLANE TRIGONOMETRY. 



In triangle BOA, 



h= V3 2 -2.5 2 

= 1.6583. 

F=ix 1.6583x5 

= 4.1458. 

sin£2)(M = £. 

log 5 = 0.69897 

colog 6 = 9.22185 - 10 

log sin iDOA = 9.92082 

iDOA= 56°26 / 33 // . 

BOA = 112° 53' 7". 

Area of sector DO J. 

112° 53' 7" „ __ 
X 28.274 



360° 
406387 



1296000 



x 28.274. 



log 406387 = 5.60894 
log 28.274 = 1.45139 
colog 1296000 = 3.88739 - 10 
log area = 0.94772 
Area sector BOA 

= 8.8658. 
Area segment BxA 
= 8.8658-4.1458 = 4.72. 
Area BACB 

= area O - [ByC + BxA] 
= 28.274 - (2.0953 + 4.72) 
= 21.4587. 

Area 224 CTO' 

= BxA - B'xC 
= 4.72-2.0953 
= 2.6247. 



Exercise XXIII. Page 90. 



1. The angle of elevation of a 
tower is 48° 19' 14" and the dis- 
tance of the base from the point of 
observation is 95 feet. Find the 
height of the tower, and the dis- 
tance of its top from the point of 
observation. 

Given A = 48° 19 / 14", b = 95 feet ; 
required a and c. 

a = b tan A. 
c = b sec A. 
log 6= 1.97772 
log tan A = 10.05045 

loga= 2.02817 
a = 106.70. 



log b : 

log sec A : 



: 1.97772 
: 0.17720 



logc = 2.15492 
c = 142.86. 



Height of tower, 106.70 feet ; dis- 
tance of top from point of observa- 
tion, 142.86 feet. 

2. From a mountain 1000 feet 
high, the angle of depression of a 
ship is 77° 35 / ll". Find the dis- 
tance of the ship from the summit 
of the mountain. 

Given B = 12° 24' 49", a = 1000 
feet ; required c. 

c = a sec B. 

log a = 3.00000 

log sec B = 0.01027 

logc = 3.01027 

c = 1023.9. 

Kequired distance, 1023.9 feet. 

3. A flagstaff 90 feet high, on a 
horizontal plane, casts a shadow of 



TEACHERS 7 EDITION. 



121 



117 feet. Find the altitude of the 
sun. 

Given a = 90 feet, b = 117 feet ; 
required A. 



tan .A = 



a 



loga = 1.95424 
colog b = 7.93181 - 10 
log tan A =9.88605 

A = 37° 34' 5". 
Altitude of sun, 37° 34' 5" '. 

4. When the moon is setting at 
any place, the angle at the moon 
subtended by the earth's radius 
passing through that place is 57' 3". 
If the earth's radius is 3956.2 miles, 
what is the moon's distance from 
the earth's centre ? 

Let C represent the place, A the 
moon, and B the earth's centre. 
Then in the right triangle ABC, 
given A = 57' 3", a = 3956.2 miles ; 
required c. 

c = a esc A. 

log a =3.59728 

log esc A = 1.78004 

logc = 5.37732 

c = 238,410. 

Moon's distance, 238,410 miles. 

5. The angle at the earth's centre 
subtended by the sun's radius is 
16' 2", and the sun's distance is 
92,400,000 miles. Find the sun's 
diameter in miles. 

Let A represent the centre of the 
earth, B that of the sun, and C a 
point on the edge of the sun's disk. 
Then in the right triangle ABC, 



given A = 16' 2", c = 92,400,000 
miles ; required 2 a. 

a = c sin A. 

logc = 7.96567 

log sin A = 7.66874 

log a = 5.63441 

a = 430,930. 

2 a = 861,860. 

Sun's diameter, 861,860 miles. 

6. The latitude of Cambridge, 
Mass., is 42° 22 x 49". What is the 
length of the radius of that parallel 
of latitude ? 




Let O be the centre of the earth, 
NS the axis, NAS the meridian of 
Cambridge, A the position of Cam- 
bridge, and C the centre of its 
parallel of latitude. Then, in the 
right triangle OAC, given O = 90° 
- 42° 22 7 49" = 47° 37 r 11", OA = 
3956.2 miles; required AC. 

AC = AO sin O. 
log AO = 3.59728 
log sin O = 9.86846 
log AC = 3.46574 

AC = 2922.4. 

Radius of parallel of latitude, 
2922.4 miles. 



122 



PLANE TRIGONOMETRY. 



7. At what latitude is the cir- 
cumference of the parallel of lati- 
tude half of that of the equator ? 

The radius of the parallel will be 
half of the radius of the earth. 

In the figure of Example 6, given 
AC — i AO ; required 90° — angle 
0, i.e., angle A. 

cos A = = ■£. 

AO 

.-. A = 60°. 
The required latitude is 60°. 

8. In a circle with a radius of 
6.7 is inscribed a regular polygon 
of thirteen sides. Find the length 
of one of its sides. 




Let be the centre of the circle, 
AB a side of the polygon, and 
the middle point. Then in the 

360° 
right triangle OCB, given O = 

= 13° 50' 46", 05 = 6.7; required 
AB = 2 CB. 

CB = OB sin BOC. 
log OB = 0.82607 
log sin 50(7 = 9.37897 
log C5 = 0.20504 
05=1.6034. 
^15 = 3.2068. 
Length of a side of the polygon, 
3.2068. 



9. A regular heptagon one side 
of which is 5.73 is inscribed in a 
circle. Find the radius of the circle. 

In the figure of Example 8, given 
BC = i x 5.73 = 2.865 and angle 

OfifiO 

BOC = - — = 25°42 / 51 // ; required 
OB. 14 

OB = BC esc BOC. 

log 50 = 0.45712 

log esc BOC = 0.36263 

log 05 = 0.81975 

OB = 6.6031. 

Radius of circle, 6.6031. 

10. A tower 93.97 feet high is 
situated on the bank of a river. 
The angle of depression of an object 
on the opposite bank is 25° 12' 54". 
Find the breadth of the river. 

Given A = 90 - 25° 12' 54" = 
64° 47' 6", b = 93.97 ; required a. 
a = b tan A. 
log 6= 1.97299 
log tan A = 10.32708 
loga= 2.30007 
a = 199.56. 
Breadth of river, 199.56 feet. 

11. From a tower 58 feet high 
the angles of depression of two 
objects situated in the same hori- 
zontal line with the base of the 
tower, and on the same side, are 
30° 13 7 18" and 45° 46' 14". Find 
the distance between these two 
objects. 

(i) Given A = 90° - 30° 13 7 18" 
= 59° 46' 42", b = 58 ; required a. 
a = 6 tan A. 



TEACHERS' EDITION. 



123 



log ?,= 1.76348 
log tan A = 10.23469 
loga= 1.99812 
a rr: 99.568. 

(ii) Given A' = 90° - 45° 46' 14" 
e= 44° 13' 46", & = 58 ; required a'. 
a' — h tan A'. 
\ogb = 1.76343 
log tan A' = 9.98832 
log a' — 1.75175 

a' = 56.461. 
a - a' = 43.107. 
Distance between the objects, 
43.107 feet. 

12. Standing directly in front of 
one corner of a flat-roofed house 
which is 150 feet in length. I 
observe that the horizontal angle 
which the length subtends has for 
its cosine Vi, and that the vertical 

angle subtended by its height has 

3 
for its sine — — What is the 

V34 
height of the house ? 

Let a = distance of observer from 
house, 
b = height of house, 
B = horizontal angle subtended 

by length of house, 
Z>' = vertical angle subtended 
by height of house. 
Then a = 150 cot B. 

b = a tan B' 
= 150 cot B tan B\ 
But cos B = V\ ; 
hence sin B = Vl — I 
2 

~V5" 



n cos B 

.-. cot B = = 1. 

sin B 

3 

Also sin T) — 

V34 



.-. cos B' — 



V34 



•. tan B' 
:e b 

Height of house, 45 feet. 



Hence b = 150 x \ x | =. 45. 



13. A regular pyramid, with a 
square base, has a lateral edge 150 
feet long, and a side of its base is 
200 feet. Find the inclination of 
the face of the pyramid to the base. 

4 




Let A be the vertex of the pyra- 
mid, BCDE its base. O the centre 
of the base, and M the middle point 
of the side BC. Required the angle. 
AMO. 

In the right triangle A OB, 
AB = 150, 

OB = ± BD = 100 V2 . 
.-. AO = VZ§ 2 _ OB 2 
= 50. 
In the right triangle A 03/, 
AO _ 50 
037" 100 
OMA = 26° 34 7 . 
Inclination of face of pyramid to 
base, 26° 34'. 



tan OMA = 



0.5. 



124 



PLANE TRIGONOMETRY. 



14. From one edge of a ditch 
36 feet wide the angle of elevation 
of a wall on the opposite edge is 
62° 39' 10". Find the length of a 
ladder that will just reach from the 
point of observation to the top of 
the wall. 

Given b = 36, A = 62° 39' 10" ; 
required c. 

c = b sec A. 

log 6 = 1.55630 

log sec A = 0.33783 

log c = 1.89413 

c = 78.367. 

Length of ladder, 78.367 feet. 

15. The top of a flagstaff has 
been broken off and touches the 
ground at a distance of 15 feet from 
the foot of the staff. If the length 
of the broken part is 39 feet, find 
the length of the whole staff. 

Given c = 39, b = 15 ; required 

c + a. 

a = V(c + b) (c - b) 



= V(39+15)(39-15) 
= V54 x 24 



= V6 2 x 9 x 4 
= 36. 
c + a = 39 + 36 = 75. 
Whole length of flagstaff, 75 feet. 

16. From a balloon, which is 
directly above one town, is observed 
the angle of depression of another 
town, 10° 14' 9". The towns being 
8 miles apart, find the height of the 
balloon. 

Given A = 90° - 10° 14' 9" = 
79° 45' 51", a = 8 ; required b. 
b = a cot A. 



log a = 0.90309 

log cot A = 9.25666 

log 6 = 0.15975 

b = 1.4446. 

• Height of balloon, 1.4446 miles. 

17. From the top of a mountain 
3 miles high the angle of depression 
of the most distant object which 
is visible on the earth's surface is 
found to be 2° 13' 50". Find the 
diameter of the earth. 




Let A be the top of the moun- 
tain, C the object observed, B the 
centre of the earth. Then given 
B = 90° - A = 2° 13' 50", AD = 3 ; 
required a. 

BC = AB cos B. 
a = (a + 3) cos B. 
.-. a (1 — cos B) — 3 cos B. 
3cos£ 



By [16], 



1 — cos B 
3 cos B 



2 sin 2 - 



log f = 0.17609 
log cos £ = 9.99967 

colog sin 2 - = 3.42154 

& 2 

log a = 3.59730 
a = 3956.4. 
2 a = 7912.8. 
Diameter of earth, 7912.8 miles. 



teachers' edition. 



125 



18. A ladder 40 feet long reaches 
a window 33 feet high on one side 
of a street. Being turned over upon 
its foot, it reaches another window 
21 feet high, on the opposite side of 
the street. Find the width of the 
street. 

Width of the one part of the street 
= V40 2 - 33 2 

=V5ii 

= 22.605. 
Width of other part 

= Vio 2 -21* 

= VTl59 

= 34.044. 
22.605 + 34.044 = 56.649. 
Total width of the street, 56.64!> 
feet. 

19. The height of a house sub- 
tends a right angle at a window on 
the other side of the street ; and the 
angle of elevation of the top of the 
house from the same point is 60°. 
The street is 30 feet wide. How 
high is the house ? 




Given CC = 30, ACC = 60°, 
BCC = 30° ; required AB. 



Now CAB = 30°. 

.: AC = 2 x CC' = 60. 

AC 

AB' 

AC 60 

cos .4 ~ ±V% 

= 40V3 = 69.282. 



cos^l 



..AB = 



Height of house, 69.282 feet. 



20. A lighthouse 54 feet high is 
situated on a rock. The angle of 
elevation of the top of the light- 
house, as observed from a ship, is 
4° 52', and the angle of elevation of 
the top of the rock is 4° 2'. Find 
the height of the rock and its dis- 
tance from the ship. 



Let 



Then 



h = height of rock. 
a = distance of ship 
h + 54 tan 4° 52' 



h 



tan 4° 2' 
tan 4° 52' 



54 

h 



h tan 4 2 
tan 4° 52' - tan 4° 2' 





tan 4° 2' 




54 




tan 4° 


2' 


tan 4° 


52'- 


tan 4° 2' 


54 


cos 4° 


52' sin 4° 2' 


sin (4 C 


52 / - 


4° 2') 


\ 1 


cos 4° 


52' sin 


4°2 r 



sin 50 r 

log 54 = 1.73239 
log cos 4° 52' = 9.99843 

log sin 4° 2' = 8.84718 

colog sin 50' = 1.83732 

log h = 2.41532 

h = 260.21. 
Also a = h cot 4° 2 / , 



126 



PLANE TRIGONOMETRY. 



\ogh= 2.41532 
log cot 4° 2' = 11.15174 
loga = 3.56706 
a = 3690.3. 
Height of rock, 260.21 feet; dis- 
tance of ship, 3690.3 feet. 

21. A man in a balloon observes 
the angle of depression of an object 
on the ground, bearing south, to be 
35° 30' ; the balloon drifts 21 miles 
east at the same height, when the 
angle of depression of the same 
object is 23° 14'. Find the height 
of the balloon. 

A' 




Can 

Let A and A' be the first and 
second positions of the balloon, re- 
spectively, C and C" the points on 
the ground directly under A and 
A\ and B the object observed. 
Then A = 54° 30', 

A' = 66° 46', 
CC = AA' = 2£. 
a = h tan ^4, 
a' = h tan ^4'. 
a' 2 - a 2 = (2*)2. 
/i 2 tan 2 ^i t= (2i)2. 
(21) 2 



A 2 tan 2 J. 

£ 2 = 



tan 2 ^4/- tan 2 J. 
Vtan 2 J/- tan 2 -4 



But tan 2 A' - tan 2 ^. 

— (tan A' + t an A) (tan J. 7 — tan J.) 

_ sin (^4/ -}- J.) sin(JL' — ^4) 



cos -4.' cos A cos -4/ cos J. 
__ sin (A' + J.) sin (A' - A) 



cos 2 A ' cos 2 ^1 



Hence 



h = 



21 cos J. 7 cos A 



Vsin (^4/ + ^.) sin (A' - A) 

log 21- 0.39794 

log cos ^. 7 = 9.59602 

log cos A = 9.76395 

colog Vsin (J/ + ^4) = 0.03408 

colog Vsin (A' - A) = 0.33636 

log h = 0.12835 

h = 1.3438. 

Height of balloon, 1.3438 miles. 

22. A man standing south of a 
tower, on the same horizontal plane, 
observes its angle of elevation to be 
54° 16' ; he goes east 100 yards, and 
then finds its angle of elevation is 
50° 8'. Find the height of the tower. 




Let AC be the tower, B and B / 
the first and second positions of the 
observer. 

Then BB' = 100. 

■a' = b cot ABG, 



TEACHERS EDITION. 



127 



a" = b cot AB'C. 
a" 2 - a' 2 = a 2 . 
b 2 (cot 2 AB'C - cot 2 ABC) = 100 2 . 

100 

o = 



Vcot 2 50° 8' - cot 2 54° W 
_ 100 sin 54° W sin 50° 8' 

Vsin 104° 24' sin 4° 8' 

log 100 = 2.00000 

log sin 54° 16' = 9.90942 

l og sin 50° 8' = 9.88510 

colog Vsin 104° 24' = 0.00693 

colog Vsin 4° 8' = 0.57110 

log b = 2.37255 

b = 235.81. 
Height of tower, 235.81 yards. 

23. The angle of elevation of a 
tower at a place A south of it is 
30° ; and at a place B, west of A, 
and at a distance a from it, the 
angle of elevation is 18°. Show 
that the height of the tower is 

— =i the tangent of 18° 

V 2 + 2 V5 

V5-1 

being ■ 



V10 + 2 V5 
With the figure and notation of 
the last example, 

n 

b 



But cot 2 1 8° = 



Vcot 2 18° -cot 2 30° 
10 + 2 V5 

6 -2 V5 
_(10 + 2V5)(6+2V5) 

6 2 - (2 V5) 2 
= 5 + 2 V5, 



and cot 2 30° 



Hence b = 



24. A pole is fixed on the top of 
a mound, and the angles of eleva- 
tion of the top and bottom of the 
pole are 60° and 30°, respectively. 
Prove that the length of the pole is 
twice the height of the mound. 



Let I 


= length of pole, 


h 


= height of mound, 


td a 


= horizontal distance of 




observer. 


Then h 


= a tan 30°. 


h + l 


= a tan 60°. 


h + l 


tan 60° 


h 


tan 30° 




= 2^ = 3. 

1 




V3 


h + l 


=z3h. 


.-. 1 


= 2h. 



V2 + 2 V5 



25. At a distance a from the foot 
of a tower, the angle of elevation A 
of the top of the tower is the com- 
plement of the angle of elevation of 
a flagstaff on top of it. Show that 
the length of the staff is 2 a cot 2 A. 

Let h = height of tower, 

and I = length of staff. 

Then h = a tan A. 

h + I = a cot A. 

I = a (cot A — tan ^1) 

cot 2 A - 1 

= a — 

cot A 

= 2acot2A. 

26. A line of true level is a line 
every point of which is equally dis- 
tant from the centre of the earth. 
A line drawn tangent to a line of 
true level at any point is a line of 



128 



PLANE TRIGONOMETRY. 



apparent level. If at any point both 
these lines are drawn,. and extended 
one mile, find the distance they are 
then apart. 




Given CA = 1 mile, BC = radius 
of the earth = 3956.2 miles ; required 
AA' = AB- CB. 

The required distance is much too 
small to be obtained by the usual 
process of solution. It is most easily 
found as follows : 

AX) 2 = AB 2 - BC 2 

= (AB-BC)(AB+BC). 



.AB-BC=- 



AC Z 



AB+ BC 

Now, as AB differs very little 
from BC, and both are very large 
in comparison with AC 2 , we may 
assume as a close approximation 
that AB = BC. Then 

AA' = AB-BC 
AC 2 



2BC 
1 



miles 



7912.4 
5280 x 12 



inches. 



7912.4 
log 5280 = 3.72263 
log 12 = 1.07918 
colog 7912.4 = 6.10169- 10 
log AA' = 0.90350 
AA' = 8.0076. 
The required distance is 8.0076 
inches. 



27. In Problem 1, page 90, determine the effect upon the computed 
height of the tower, of an error in either the angle of elevation or the 
measured distance. 

With the notation of Problem 1, suppose that the error in the angle is 
e\ and that in the measured distance is e 2 . Then the formulas 

a = 6tan^i, c = 6sec^4 

become a = (b + e 2 ) tan (A + ei), c = (b -f e 2 ) sec (^4 + ei), 

and the error in the computed value of a is 
(b + e 2 ) tan (A -f e{) — b tan A 

— b {tan (A + ei) — tan A} + e 2 tan (A + ei) 
b sin e\ 



cos {A -f e{) cos A 
or, approximately, for small errors, 

bet 



+ e 2 tan (A + ei), 



cos 2 A 
where e\ is measured in radians. 



+ e 2 tan A, 



TEACHERS EDITION. 



129 



+ e>2 sec (A -f e{) 



The error iu c is 

b [sec (A -f ei) - sec A) 4 e 2 sec (A + ei) 
6 [cos A — cos (^1 -f e\) 
cos (^4 + ei)cosJ. 

2 6 sin M + i ei) sin (i ei) , ,a,\ 

By [23], = \ , 1; f-^ + e 2 sec (J. + Ci), 

cos (^4 + ei)cos^l 

or, approximately, for small errors, 

be 1 sin ^4 



cos 2 ^1 



4- e 2 sec .4 = (hex tan ^1 4- e 2 ) sec A. 



28. To determine the height of 
an inaccessible object situated on 
a horizontal plane, by observing its 
angles of elevation at two points 
in the same line with its base, and 
measuring the distance between 
these two points. 
A 




Let AC be the object, B and B' 
the two points of observation. Then 
given the angles B' and ABC, and 
the side BB' ; required A C. 



AB = BB' 



= BB 



sin BAR 
sin & 



sin (A BC - B') 
AC = AB sin ABC 

_ sin B' sin ABC 
sin(ABC -B')' 

29. The angle of elevation of an 
inaccessible tower situated on a 
horizontal plane is 63° 26' ; at a 
point 500 feet farther from the base 
of the tower the angle of eleva- 
tion of its top is 32° 14'. Find the 
height of the tower. 



From the solution of Example 28, 
k sin 32° 14' sin 63° 26' 



AC = 500 - 



500 



sin (63° 26' - 32° 14') 

sin 32° 14' sin 63° 26' 

sin 31° 12' 



log 500 = 2.69897 

log sin 32° 14' = 9.72703 

log sin 63° 26' = 9.95154 

colog sin 31° 12' = 0.28565 

log ^10 = 2.66319 

AC = 460.46. 

Height of the tower, 460.46 feet. 

30. A tower is situated on the 
bank of a river. From the oppo- 
site bank the angle of elevation of 
the tow r er is 60° 13', and from a 
point 40 feet more distant the angle 
of elevation is 50° 19'. Find the 
breadth of the river. 

In the figure for the solution of 
Example 28. 

CB = AB cos ABC 

sin B' cos ABC 
= BB 



Hence 
C£ = 40 



sin (ABC - B*) 

sin 50° 19' cos 60° 13' 
sin 9° 54' 



130 



PLANE TRIGONOMETRY. 






log 40 = 1.60206 

log sin 50° 19' = 9.88626 

log cos 60° 13' = 9.69611 

colog sin 9° 54' = 0.76465 

log CB = 1.94908 

CB = 88.936. 

Breadth of river, 88.936 feet. 

31. A ship sailing north sees two 
lighthouses 8 miles apart, in a line 
due west ; after an hour's sailing, 
one lighthouse bears S.W., and the 
other S.S.W. Find the ship's rate. 

In the figure for the solution of 
Example 28, let B and B' be the 
lighthouses, C the original position 
of the ship, and A its final position. 
Then CM £=22° 30' and CAR =46°; 
hence ABC = 67° 30' and B' = 45°. 

a n _ q sin 45 ° sin 67 ° 80/ 
A O — o 

sin 22° 30' 
= 8 sin 45° cot 22° 30'. 
log 8= 0.90309 
log sin 45°= 9.84949 
log cot 22° 30' = 10.38278 
log^C = 1.13536 
^10 = 13.657. 
Ship's rate, 13.657 miles per hour. 

32. To determine the height of 
an accessible object situated on an 
inclined plane. 

A 




Let CBB' be the inclined plane, 
AC the object, B and B' two points 
of observation, AC' the perpendic- 



ular from A on CBB'. Then given 
CB, BB', and the angles ABC, B'; 
required AC. 

From the solution of Example 28, 
„ sin B' sin ABC 



AC = BB' 



and 



CB = BB 



sin (ABC - B') 
sin jB' cos ABC 



and 



sin (ABC - B') 
Then CC = CB - CB, 



AC: 



LC' 2 + CC 1 . 



33. At a distance of 40 feet from 
the foot of a tower on an inclined 
plane the tower subtends an angle 
of 41° 19' ; at a point 60 feet farther 
away the angle subtended by the 
tower is 23° 45'. Find the height 
of the tower. 

From the solution of Example 32, 

A nr «A Sm 23 ° 45 ' Sil1 41 ° 19 ' 



CB = 60 



sin 17° 34' 
sin23°45 / cos41°19' 



sin 17° 34' 

log 60 = 1.77815 

log sin 23° 45' = 9.60503 

log sin 41° 19' = 9.81969 

colog sin 17° 34' = 0.52026 

log^lC 7 = 1.72313 

AC = 52.860. 

log 60 = 1.77815 

log sin 23° 45' = 9.60503 

log cos 41° 19' = 9.87568 

colog sin 17° 34' = 0.52026 

log C'B= 1.77912 

C"J5 = 60.134. 

CC = 60.134 -40 

= 20.134. 

AC 



tan ACC = 



CC 



TEACHERS EDITION. 



131 



AC — A C esc A CC. 
logvlC" = 1.72313 
colog C'C = 8.69607 - 10 
log tan ACC = 0.41920 

4(7(7 = 69° 8' 55". 

log AC' = 1.72313 

log esc ACC = 0.02941 

log AC- 1.75254 

AC = 56.564. 

Height of tower, 56. 564 feet. 

34. A tower makes an angle of 
113° 12' with the inclined plane on 
which it stands ; and at & distance 
of 89 feet from its base, measured 
down the plane, the angle subtended 
by the tower is 23° 27'. Find the 
height of the tower. 

In the triangle ACB, given CB = 
89 feet, C = 113° 12', B = 23° 27' ; 
required AC. 

A = 180° - (B + C) 
= 43° 21'. 

AC=CB*™*. 

sin A 

log 89 = 1.94939 

logsin23°27 / = 9.59983 

colog sin 43° 21' = (U6330 

log^C= 1.71261 

AC = 51.595. 

Height of tower, 51. 595 feet. 

35. From the top of a house 42 
feet high the angle of elevation of 
the top of a pole is 14° 13' ; at the 
bottom of the house it is 23° 19'. 
Find the height of the pole. 

Let A be the top of the pole, B 
and B' the top and bottom of the 
house, and C the foot of .the per- 



pendicular from A on BB' ; re- 
quired B'C. 

From the solutions of Examples 
28 and 30. 

sin A B'C cos ABC 



CB = BB' 



42 



sin(ABC -AB'C), 
sin 66° 41' cos 75° 47' 



sin 9° 6' 

log 42 = 1.62325 
log sin 66° 41' = 9.96300 
log cos 75° 47' = 9.39021 
colog sin 9° (Y = 0.80091 
log CB = 1.77737 
CB = 59.892. 
B'C = CB + BB' 
= 59.892 +42 
= 101.892. 
Height of pole, 101.892 feet. 

36. The sides of a triangle are 
17, 21, 28. Prove that the length 
of a line bisecting the greatest side 
and drawn from the opposite angle 
is 13. 

Let a = 28, b = 21, c = 17, 
Then [26] 
17 2 = 28 2 + 2V 2 - 2 x 28 x 21 cos C; 
to prove that 

132= 14 2 -f2r 2 -2 x 14x21 cos C. 
Subtract the first equation from 
twice the second, 

2 x 13 2 - 17' 2 = 2 x U' 2 - 28* + 21 2 
= 21 2 - 2 x 14 2 
= 7 2 (3 2 - 2 s ) = 7 2 . 
2 x 169 - 289 = 49, 
49 = 49. 

37. A privateer 10 miles S.W. of 
a harbor sees a ship sail from it in 
a direction S. 80° E. at a rate of 



132 



PLANE TRIGONOMETRY. 



9 miles an hour. In what direction, 
and at what rate, must the privateer 
sail in order to come up with the 
ship in 1-J hours ? 

Let A be the harbor, B the 
original position of the privateer, 
and C the point where the vessels 
are to meet. Then ^. = 125°, 6 = 131, 



c = 10 ; required B and 



1* 



t2Llii(B- C) = ; 



-tan i(B + C) 



-tan 27° 30' 



6 + c 
_ 3.5 
~23.5" 
= ? V tan 27° 30'. 
log 7 = 0.84510 
colog47 = 8.32790 - 10 
logtan27°30' = 9.71648 



log 


tani(£- 


-C) = 


8.88948 






i(B- 


-C) = 


4° 26'. 






B- 


- c = 


8° 52'. 






B+ C = 


55°. 








•. B = 


31° 56'. 








a = 


.sin A 

o 

sin B 











13 fi Sin 


125° 



sin 31° 56' 

log 13.5 = 1.13033 

log sin 125° = 9.91336 

colog sin 31° 56' = 0.27660 

loga = 1.32029 

a = 20.907. 

a 

H 

Privateer's course, 31° 56' E. of 
N.E., or N. 76° W E.; rate 13.938 
miles per hour. 

38. A person goes 70 yards up a 
slope of 1 in 3£ from the edge of 



13.938. 



a river, and observes the angle 
of depression of an object on the 
opposite bank to be 2i°. Find 
the breadth of the river. 

Let A and B be the original and 
final positions of the observer, and 
C the object observed. Then given 
c = 70, C = 2i°, A = 180° - 

tan— 1 — ; required b. 



A 


= 180° - 


tan- i f 




= 180° - 


tan- 10.2857 




= 180° - 


15° 56' 40" 




= 164° 3' 


20". 


B 

b 


= 180° - 
= 13° 4r 

sin B 

= c 


[A + 6) 

40". 



sin C 

log 70 = 1.84510 

log sin 13° 4r 40" = 9.37428 

colog sin 2° 15 7 = 1.40605 

log b = 2.62543 

6 = 422.11. 

Breadth of river, 422.11 yards. 

39. The length of a lake sub- 
tends, at a certain point, an angle 
of 46° 24 7 , and the distances from 
this point to the two extremities of 
the lake are 346 and 290 feet. Find 
the length of the lake. 

Given A = 46° 24 7 , 6=346, c=290; 
required a. 

tan i(B - O) = ^— - tan i(B + C) 
6 + c 

56 

636 

log 56= 1.74819 

colog 636= 7.19654- 

log tan m° 48' = 10.36795 

log tani (B - C) = 9.31268 



= — tan 66° 48'. 



10 



TEACHERS' EDITION. 



133 



$(B-C)= 11° 36' 33". 

B - C = 23° 13' 6". 

B + C = 133° 36'. 

.-. J3 = 78° 24' 33". 

sin^L ,_ sin 46° 24' 
a = b - — - = 346 



' sin B ~ sin 78° 24' 33" 

log 346 = 2.53908 

log sin 46° 24' = 9.85984 

colog sin 78° 24' 33" = 0.00895 

log a = 2.40787 

a = 255.78. 

Length of lake, 255.78 feet. 

40. Two ships are a mile apart. 
The angular distance of the first 
ship from a fort on shore, as ob- 
served from the second ship, is 
35° 14' 10"; the angular distance 
of the second ship from the fort, 
observed from the first ship, is 
42° 11' 53". Find the distance in 
feet from each ship to the fort. 

Given B = 35° 14' 10", C = 
42° 11' 53", a = 5280 ; required b 
and c. 

A = 180-(E + C) 

= 102° 33' 57". 

sin B 

b = a 

sin A 

log 5280 = 3.72263 

log sin 35° 14' 10" = 9.76114 

colog sin 102° 33' 57" = 0.01053 

log b = 3.49430 

6 = 3121.1. 

sin C 

c = a 

sin^i 

log 5280 = 3.72263 

log sin 42° 11' 53" = 9.82717 

colog sin 102° 33' 57" = 0.01053 

logc = 3.56033 



c = 3633.5. 

Distance of first ship from fort, 
3121.1 feet; of second ship from 
fort, 3633.5 feet. 

41. Along the bank of a river is 
drawn a base line of 500 feet. The 
angular distance of one end of this 
line from an object on the opposite 
side of the river, as observed from 
the other end of the line, is 53° ; 
that of the second extremity from 
the same object, observed at the 
first, is 79° 12'. Find the breadth 
of the river. 

Given £=53°, C = 79°12', a = 500; 
required p, the perpendicular from 
A on a. 

A = 180° - (B + C) 
= 47° 48'. 
sin B 



b = a- 



sin^ 



j) = b sin C = a 
= 500 



sin B sin C 



sin A 
sin 53° sin 79° 12' 



sin 47° 48' 



log 500 = 2.69897 

log sin 53° = 9.90235 

log sin 79° 12' = 9.99224 

colog sin 47° 48' = 0.13030 

logp = 2.72386 

p = 529.49. 
Breadth of river, 529.49 feet. 

42. A vertical tower stands on a 
declivity inclined 15° to the horizon. 
A man ascends the declivity 80 feet 
from the base of the tower, and 
finds the angle then subtended by 



134 



PLANE TRIGONOMETRY. 



the tower to be 30°. Find the 
height of the tower. 

Let A and B be the top and bot- 
tom of the tower, and C the position 
of observation. Then given a = 80, 
B = 75°, C = 30° ; required c. 
A = 180° - (B + C) = 75°. 

a sin C 

c = 

sin A 

80 sin 30° 



sin 75° 

log 80 = 1.90309 

log sin 30° = 9.69897 

colog sin 75° = 0.01506 

logc = 1.61712 

c = 41.411. 

Height of tower, 41.411 feet. 

43. The angle subtended by a 
tower on an inclined plane is, at 
a certain point, 42° 17 / ; 325 feet 
farther down, it is 21° 47'. The 
inclination of the plane is 8° 53'. 
Find the height of the tower. 

A 



AB = BB' 




sin BAB' 
sin B' 



AC = 



BB' 

sin(B-B') 

AB sin B 



sin C 



= BB' - 



sin B sin B' 



325 



sin C sin (B - B') 
sin 42° 17 / sin21°47 / 
sin 98° 53' sin 20° 30' ' 






log 325 = 2.51188 

log sin 42° 17' = 9.82788 

log sin 21° 47' = 9.56949 

colog sin 98° 53' = 0.00524 

colog sin 20° 30' = 0.45567 

log AC = 2.37016 

AC = 234.51. 

Height of tower, 234.51 feet. 

44. A cape bears north by east, 
as seen from a ship. The ship sails 
northwest 30 miles, and then the 
cape bears east. How far is it from 
the second point of observation ? 

Let A be the cape, B and C the 
first and second positions of the 
ship. Then given B = 56° 15', 
C = 45°, a = 30 ; required b. 

A = 180° - (B + C) = 78° 45 7 . 

. a sin B 30 sin 56° 15' 

b = = 

sin A sin 78° 45 / 

log 30 = 1.47712 

log sin 56° 15' = 9.91985 

colbg sin 78° 45' = 0.00843 

log b = 1.40540 

b = 25.433. 

Distance of cape from second 

point of observation, 25.433 miles. 

45. Two observers, stationed on 
opposite sides of a cloud, observe 
its angles of elevation to be 44° 56 7 
and 36° 4 7 . Their distance from 
each other is 700 feet. What is the 
height of the cloud ? 

Given A = 44° 56^ B = 36° 4', 
c = 700 ; required the perpendicular 
p from C on c. 

C=l$0° -(A + B) =99°. 
sin B 



bt=c- 



sinc 



TEACHERS EDITION. 



135 



p — b sin A = c 



sin B sin A 



sin C 

„, r sin 36° 4' sin 44° 50' 

= /GO 

sin 99° 

log 700 = 2.84510 

log sin 36° 4' = 9.76991 

log sin 44° 56' = 9.84898 

colog sin 99° = 0.00538 

logp = 2.46937 

p = 294.69. 

Height of cloud, 294.69 feet. 

46. From a point B at the foot 
of a mountain, the angle of eleva- 
tion of the top A is 60°. After 
ascending the mountain one mile, 
at an inclination of 30° to the hori- 
zon, and reaching a point C, the 
angle ACB is found to be 135°. 
Find the height of the mountain in 
feet. 
A 




CD = CB sin CBD 
= 5280 x i = 2640. 
CB sin CB A 
sin CAB 
AE = AC sin EC A 

_ CB sin CBA sin EC A 
sin CAB 
5280 sin 30° sin 75° 



sin 15° 

5280 xicosl5° 



sin 15° 
= 2640 cot 15°. 



log 2640= 3.42160 

log cot 15° = 10.57195 

\ogAE = 3.99355 

AE = 9852.6. 
AF= AE + CI) 
= 12,492.6. 
Height of the mountain, 12.492.6 
feet. 

47. From a ship two rocks are 
seen in the same right line ^Yith the 
ship, bearing N. 15° E. After the 
ship has sailed northwest 5 miles, 
the first rock bears east, and the 
second northeast. Find the dis- 
tance between the rocks. 




Let A and B be the two rocks, 
C and C the first and second posi- 
tions of the ship. Then given 
C = 60°, CC'B = 45°, CCA = 90°, 
CC = 5 ; required AB. 

AC= CC' sec C = 5x 2 = 10. 



BC = CC' 



sin BC'C 



, sin 45 c 



sin CBC sin 75° 

log 5 = 0.69897 

log sin 45° = 9,84949 

colog sin 75° = 0.01506 

log BC = 0:56352 

BC = 3.6603. 



136 



PLANE TRIGONOMETRY. 



AB = AC-BC 
= 6.3397. 
Distance between rocks, 6.3397 
miles. 

48. From a window on a level 
with the bottom of a steeple the 
angle of elevation of the steeple is 
40°, and from a second window 18 
feet higher the angle of elevation 
is 37° 30'. Find the height of the 
steeple. 

Let A and B be the windows, 
and C the top of the steeple. Then 
given c = 18, A = 50°, B = 127° 30'; 
required height of steeple. 

h = b sin 40°. 

C = 180° - (A + B) = 2° 30'. 



sin B 



18 



sin 127° 30' 



ft =18 



sin C sin 2° 30' 

sin 127° 30' sin 40° 



sin 2° 30' 



log 18 = 1.25527 

log sin 127° 30' = 9.89947 

log sin 40° = 9.80807 

colog sin 2° 30' = 1.36032 

log h = 2.32313 

h = 210.44. 

Height of steeple, 210.44 feet. 

49. To determine the distance 
between two inaccessible objects by 
observing angles at the extremities 
of a line of known length. 




Let A and B be the inaccessible 
objects, C and D the extremities of 
the given line. Then, given CD, 
ACD, BCD, ADC, and BDC; re- 
quired A B. 



AC= CD 



BC = CD 



sin ADC 

sin CAD ' 

sin BDC 



sin CBD 
Then, in the triangle CAB, two 
sides and the included angle are 
known, and the third side can be 
computed as usual. 

50. Wishing to determine the 
distance between a church A and a 
tower B, on the opposite side of a 
river, I measure a line CD along the 
river (C being nearly opposite A), 
and observe the angles A CB, 58° 20'; 
ACD, 95° 20'; ADB, 53° 30'; BDC, 
98° 45'. CD is 600 feet. What is 
the distance required ? 

From the solution of Example 49, 

AC=CD sinABC 

sin CAD 

sin 45° 15' 

sin 39° 25' ' 

sin BDC 

sin CBD 

rin98'4r 

sin 44° 15' 

log 600 = 2.77815 

log sin 45° 15' = 9.85137 

colog sin 39° 25' = 0.19726 

log AC = 2.82678 

AC = 671.09. 

log 600 = 2.77815 

log sin 98° 45' = 9.99492 

colog sin 44° 15' = 0.15627 

log £0 = 2.92934 



600 



BC= CD 



TEACHERS EDITIOX. 



137 



BC = 849.84. 

tan i (CAB- CBA) 

BC-AC CAB CBA) 

BC + AC v 

= -AI^i t an60°50'. 
1520.93 

log 178.75= 2.25224 
colog 1520.93 = 6.81789-10 
log tan 60° 50' = 10.25327 
logtani((M£-C7L4) 

= 9.32340 

i (CAB - CBA) = 11° 53' 28" 
i (CAB + CBA) = 60° 50' 

CAB = 72° 43' 28" 



= 440 



sin 67° 14' 



AB=BC 



; 849.84 



sin AC B 
sin CAB 

sin 58° 20' 



sin 72° 43' 28" 



log 849.84 = 2.92934 

log sin 58° 20' = 9.92999 

colog sin 72° 43' 28" = 0.02005 

log^.£ = 2.87938 

AB = 757.50. 
Required distance, 757.50 feet. 

51. Wishing to find the height of 
a summit A, I measure a horizontal 
base line CD, 440 yards. At C, the 
angle of elevation of A is 37° 18', 
and the horizontal angle between D 
and the summit is 76° 18' ; at D the 
horizontal angle between C and the 
summit is 67° 14'. Find the height. 

Let A' be the point directly under 
^4, in the same horizontal plane 
with CD. Then, in the triangle 
A'CD, 



A'C = CD 



sinD 
sin A' 



sin 36° 28' 
AA' = A'C tan AC A' 

= 440 Sin67 ° 14/ tan37oi8- 
sin 36° 28' 

log 440 = 2.64345 

log sin 67° 14' = 9.96477 

log tan 37° 18' = 9.88184 

colog sin 36° 28' = 0.22595 

log AA' = 2. 71601 

AA' = 520.01. 

Height, 520.01 yards. 

52. A balloon is observed from 
two stations 3000 feet apart. At 
the first station the horizontal angle 
of the balloon and the other station 
is 75° 25', and the angle of elevation 
of the balloon is 18°. The horizon- 
tal angle of the first station and the 
balloon, measured at the second 
station, is 64° 30'. Find the height 
of the balloon. 

Let B be the first station, C the 
second, A the position of the balloon, 
and A' the point directly under A, 
in the same horizontal plane as BC. 
Then 

AA' -= A' B tan A' B A 
sinA'CB^ 



BC- 



= 3000 



sin BA'C 
sin 64° 30' 



tan A'BA 



tan 18° 



sin 40° 5' 

log 3000 = 3.47712 

log sin 64° 30' = 9.95549 

log tan 18° = 9.51178 

colog sin 40° 5' = 0.19118 

log A A' = 3.13557 

AA' = 1366.4. 

Height of balloon, 1366.4 feet. 



138 



PLANE TRIGONOMETRY, 



53. Two forces, one of 410 
pounds, and the other of 320 
pounds, make an angle of 51° 37'. 
Find the intensity and the direction 
of their resultant. 




Let AB and AD represent the 
forces, and AG their resultant. 
Then, in the triangle ABC, given 
c = 410, a = 320, B = 180° - 51° 37' 
= 128° 23' ; required b and A. 

tan |(C - A) 
c — a 



c -f a 
90 



tani(0-f A) 



- tan 25° 48' 30". 
730 

log 90 = 1.95424 

log tan 25° 48" 30" = 9.68448 

colog 730 = 7.13668-10 



tan-i(C-^i) = 8.77540 
i(G-A)= 3°24 / 43 // 
j(C + A) = 2b° 48' SO" 



A = 22° 23 / 47 A 



b = a 



sin B 



= 320 



sin^L 

sin 128° 23' 



sin 22° 23" 47" 
log 320 = 2.50515 

log sin 128° 23' =9.89425 
colog sin 22° 23' 47" = 0.41906 
log b = 2.81846 
b = 658.36. 
Intensity of resultant, 658.36 
pounds ; angle between resultant 
and first force, 22° 23' 47". 



54. An unknown force, combined 
with one of 128 pounds, produces a 
resultant of 200 pounds, and this 
resultant makes an angle of 18° 24' 
with the known force. Find the 
intensity and direction of the un- 
known force. 

In the figure for the solution of 
Example 53, given, in the triangle 
ABC, c = 128, A = 18° 24', b = 200 ; 
required a and B. 

b-c 



tan* (J?- C): 



'■ tan I ( B + C) 



= — tan 80° 48'. 



b + c 

328 

log 72= 1.85733 
log tan 80° 48' = 10.79058 
colog 328 = 7.48413 - 
log tail i(JB- C) = 10.13204 

i(B-C)= 53° 34' 44' 

i(B + C) = JXF4W__ 

B = 134° 22' 44' 

180° - B = 45° 37' 16". 



10 



b sin A 



= 200- 



sinl8°24' 



sin B sin 134° 22 / 44" 

log 200 = 2.30103 
log sin 18° 24/ = 9.49920 
colog sin 134° 22' 44" = 0.14586 
loga = 1.94609 
a = 88.326. 
Intensity of unknown force, 88.326 
pounds ; angle between known and 
unknown forces, 45° 37' 16". 

55. At two stations, the height 
of a kite subtends the same angle A. 
The angle which the line joining 
one station and the kite subtends 
at the other station is B ; and the 
distance between the two stations 



TEACHERS' EDITION. 



139 



is a. Show that the height of the 
kite is i a sin A sec B. 

Let C be the position of the kite, 
D and E the stations, and C the 
point directly under C in the same 
horizontal plane with BE. 

Since the elevation of the kite is 
the same at B and E, the triangle 
CDE is isosceles, and 

CB = CE = iasecB. 

Also CC = CD sin A 

= i a sin A sec B. 

56. Two towers on a horizontal 
plane are 120 feet apart. A person 
standing successively at their bases 
observes that the angle of elevation 
of one is double that of the other ; 
but, when he is halfway between 
them, the angles of elevation are 
complementary. Prove that the 
heights of the towers are 90 and 
40 feet. 

Let A and B be the tops of the 
towers, A' and B' their bases, and 
C the point halfway between them. 
Then the triangles AA'C and BB'C 
are similar, and 

AA' _A'C 
~RC~~BB'' 
AA' x BR = B'C x A'C 
= 3600. 
Also AB'A' = 2 BA'B'. 

2 tan BA'B' 



.-. tan AB'A' ■. 

AA' 
120 



1 - tan 2 BA'B' 
n BB' 



120 



BB' Z 
120 2 " 



240 BW 

" 120 2 - BB' 2 ' 

AA'(120 2 -BB' 2 ) = 120 x 240 BB'. 

^?(120 2 -RB' 2 ) = 120 x 240 BB'. 
BB' 

120 2 - BR 2 = 8 BR 2 . 

BR 2 = 40 2 . 

BB' = 40. 

AA' = 90. 

Heights of towers, 90 feet and 40 

feet. 

57. To find the distance of an 
inaccessible point C from either of 
two points A and B, having no 
instruments to measure angles. 
Prolong CA to a, and CB to 6, 
and join AB, Ab, and Ba. Measure 
AB, 500; aA, 100; aB, 560; bB, 
100 ; and Ab, 550. Compute the 
distances AC and BC. 




In the triangle aAB, 
s = ^-(500 + 100 + 560): 



580. 



•4 



'80x480 



10 



tan 4- aAB 

? \580x20 

log 96= 1.98227 

colog 29 = 8.53760 

2 ) 0.51987 

log tan £ aAB = 10.25993 

iaAB= 61°12 / 20 // . 

aAB = 122° 24 / 40". 

CAB= 57°35 / 20". 

In the triangle bAB, 
s = i (500 + 550 + 100) = 575. 



140 



PLANE TRIGONOMETRY. 



tan^2M = J 75x475 = J 
\ 575x25 \2 

log 57= 1.75587 
colog 23 = 8.63827-10 
2 ) 0.39414 
log tani bBA = 10.19707 

ibBA= 57°34 / 30 // . 
bBA = 115° 9', 
CBA = 64° 51'. 
In the triangle ABC, 

A = 57° 35 / 20", 
B = 64° 51', 
C = 57° 33' 40". 
.sin A 



'67 ; 
23 



BC = AB 



:500 



AC = AB 



= 500 



sin O 
sin 57° 35 / 20" 
sin 57° 33" 40' 
sin B 
sin (7 
sin 64° 51' 



sin 57° 33' 40" 

log 500 = 2.69897 

log sin 57° 35' 20" = 9.92646 

colog sin 57° 33' 40" = 0.07368 

log BC = 2.69911 

BC = 500.16. 

log 500 = 2.69897 

log sin 64° 51' = 9.95674 

colog sin 57° 33' 40" = 0.07368 

log AC = 2.72939 

AC = 536.28. 

Distances of C from ^4 and B, 

536.28 feet; 500.16 feet. 

58. Two inaccessible points A 
and B are visible from D, but no 
other point can be found whence 
both are visible. Take some point 
C whence A and D can be seen, 



and measure CD, 200 feet ; ADC, 
89°; ACD, 50° 30'. Then take 
some point E whence D and B are 
visible, and measure DE, 200 feet ; 
BDE, 54° 30'; BED, 88° 30'. At 
D measure ADB, 72° 30'. Compute 
the distance AB. 




AD= CD 



sin A CD 
sin CAD 

= 200 Sin50 ° 8(r . 
sin 40° 30' 

log 200 = 2.30103 

log sin 50° 30' = 9.88741 

colog sin 40° 30' = 0.18746 

log AD = 2.37590 

AD = 237.63. 
= 200 



sin DBE 

sin 88° 30' 



sin 37° 

log 200 = 2.30103 

log sin 88° 30' = 9.99985 

colog sin 37° = 0.22054 

log BD = 2.52142 

BD = 332.22. 
tan i (DAB - DBA) 
BD-AD . 



BD + AD 
94.59 



tsml(DAB + DBA) 



569.85 



tan 53° 45'. 



teachers' edition 



141 



log 94.59= 1.97585 

colog 569.85= 7.24424 

log tan 53° 45' = 10.13476 

log tan $ (DAB - DBA) = 9.35485 

i (DAB - DBA) = 12° 45' 21" 

i (DAB + DBA) = 53° 45" 

DAB = 66° 30' 21" 
smADB 



AB = BD 



= 332.22 



sin DAB 

sin 72° 30' 



sin 06° 30' 21" 

log 332. 22 = 2.52142 

log sin 72° 30' = 9.97942 

colog sin ffi° 30' 21" = 0.03758 

\ogAB = 2.53842 

^15 = 345.48. 
Distance ^1£, 345.48 feet. 

59. To compute the horizontal 
distance between two inaccessible 
points A and J5, when no point can 
be found whence both can be seen. 
Take two points C and D, distant 
200 yards, so that A can be seen 
from C, and B from D. From C 
measure CF, 200 yards to F, whence 
A can be seen ; and from D, meas- 
ure DE, 200 yards to E, whence B 
can be seen. Measure AFC y 83° ; 
ACD, 53°30 / ; ACF, 54° 31'; JSZ^, 
54° 30' ; BDC, 156° 25' ; DEB, 
88° 30'. 




AC=CF 



sin (LI 2^ 



= 200 



sin 83° 



sin 42° 29' 

log 200 = 2.30103 

log sin 83° = 9.99675 

colog sin 42° 29' = 0.17045 

log AC = 2.46823 

AC = 293.92. 

, sin BED 



BD = DE 



= 200 



sin DM? 

sin 88° 30' 



sin 37° 

log 200 = 2.30103 

log sin 88° 30' = 9.99985 

colog sin 37° = 0.22054 

log BD = 2.52142 

BD = 332.22. 

t2mi(ADC-CAD) 

AC — CD 

- tan $ (ADC + CAD) 



AC + CD 
93.92 



tan 63° 15'. 
493.92 

log 93. 92= 1.97276 
colog 493. 92 = 7. 30634 - 1 
log tan 63° 15' = 10.29753 
\ogt2Lii$(ADC-CAD) 

= 9.57663. 

i(ADC - CAD) = 20° 40 / 8" 
i (ADC + CAD) = 63° 15' 

ADC = 83° 55 / 8" 



AD = AC 



= 293.92 



sin A CD 
sin ADC 

sin 53° 30' 



sin 83° 55' 8" 

log 293. 92 = 2.46823 

log sin 53° 30' = 9.90518 

colog sin 83° 55' 8" = 0.00245 

log AD = 2.37586 

^LD = 237.61. 



142 



PLANE TRIGONOMETRY. 



BBA = BBC - ABC 

= 156° 25' - 83° 55' 8" 
= 72° 29' 52". 

ttmi(BAB-BBA) 

BB-AB i, DAB + B BA) 
BB + AB 2K ' 

= J^Itan53°45'4". 
569.83 

log 94.61 = 1.97594 
colog 569.83 = 7.24426 - 10 
log tan 53° 45' 4" = 10.13478 
log t2Lni(BAB-BBA) 

= 9.35498 

i (BAB - BBA) = 12° 45' 35" 
i (BAB + BBA) = 53° 45' 4" 
BAB = 66° 30' 39" 
v sin ABB 



AB = BB- 



332.22 



sin BAB 

sin 72° 29' 52" 



sin 66° 30' 39" 

log 332.22 =2.52142 

log sin 72° 29' 52" = 9.97941 

colog sin 66° 30' 39" = 0.03757 

log AB = 2.53840 

AB = 345.46. 

Distance AB, 345.46 yards. 

60. A column in the north tem- 
perate zone is east-southeast of an 
observer, and at noon the extremity 
of its shadow is northeast of him. 
The shadow is 80 feet in length, and 
the elevation of the column, at the 
observer's station, is 45°. Find the 
height of the column. 

Let A be the observer's position, 
B the extremity of the shadow, and 
C the base of the column. Then 
given A = 67° 30', C = 67° 30', 
a = 80 ; ■ required b. 



b = a 



sin B 



80 



sin A 
sin 45° 



sin 67° 30' 

log 80 = 1.90309 

log sin 45° = 9.84949 

colog sin 67° 30' = 0.03438 



log 6= 1.7* 
b = 61.23. 
Let B' be the top of the col- 
umn. Then A AB' C is isosceles 
since A = B' = 45°. 

Therefore, the height of the 
column is 61.23 feet. 

61. From the top of a hill the 
angles of depression of two objects 
situated in the horizontal plane of 
the base of the hill are 45° and 30° ; 
and the horizontal angle between 
the two objects is 30°. Show that 
the height of the hill is equal to the 
distance between the objects. 

Let A be the top of the hill, A' 
the point directly under A in the 
horizontal plane of the base of the 
hill, B and C the objects observed. 

Then 

A'B = A' A. 

A'C = A' A tan 60° = VsA'A. 

BC 2 = 2FJB 2 + WC 1 

- 2 A'B x A'C cos BA'C 
= A 7 A 2 + 3A 7 A 2 

-2 A' A x V3 A' A x i V3 
= AA 2 + 3Z 7 Z 2 - 3 ATA 2 



= A'A\ 
BC = A' A. 

62. Wishing to know the breadth 
of a river from A to B, I take A C, 



TEACHERS 7 EDITION. 



143 



100 yards in the prolongation of 
BA, and then take CD, 200 yards 
at right angles to AC. The angle 
BDA is 37° 18' 30". Find AB. 




tan ADC = 



AC 100 



CD 200 2 
log tan ADC = 9.69897. 

ADC = 26° 33' 54". 
5DC = ^1D5 + -4£>C 
= 63° 52' 24". 
5C = ODtan5Z)C 

= 200 tan 63° 52' 24". 
log 200= 2.30103 
log tan 63° 52' 24" = 10.30939 
log BC = 2.61042 
5(7 = 407.77. 
^LB = 5C-^iC 

= 307.77. 
.45 = 307.77 yards. 

63. The sum of the sides of a 
triangle is 100. The angle at A is 
double that at B, and the angle at 
B is double that at C. Determine 
the sides. 

5 = 2 C. 
A=2B = ±C. 
A+B+C=7C= 180°. 

.-. C = 25° 42 / 51f". 
5= 61°25'42f". 
A = 102° 5r 25f". 
a _ sin ^4 
c sin C 



log sin A =9.98897 
cologsin C = 0.36263 



, a 
log- 


= 0.35160 


a 

c 


= 2.247. 


a = 


= 2.247 c. 


6 


sin B 


c 


sin C 



log sin 


B = 


9.89311 


colog sin 


C = 


0.36263 



log- = 0.25574 



= 1.802. 



b= 1.802 c. 
a + b + c = (2.247 + 1.802 + l)c 
= 5.049 c. 

., C= J«L = 19 . 806 

5.049 

a = 2.247c= 44.504 

6 = 1.802 c = 35.690 

a + 6 + c = 100.000 

The sides are 19.8, 35.7, 44.5. 

64. If sin2 ^1 + 5 cos 2 ^4 = 3, 
find A. 

sin" 2 A + 5 cos 2 A = 3. 

sin 2 ^L + 5 - 5 sin 2 J. = 3. 

4 sin 2 ^4 = 2. 

sin 2 A = i. 

sinA=± y/%. 

.*. A = ± 45°, ± 135°. 

65. If sin 2 J. = m cos ^1 — n, 
find cos A. 

sin 2 ^4 = m cos ^4 — n . 
1 — cos 2 A = m cos A — n. 
cos 2 .A 4- in cos ^4 = n + 1. 
4 cos 2 A + ( ) + w 2 = "i 2 + 4 (n + 1). 



144 



PLANE TRIGONOMETRY. 



2 COS A + m 



:±Vm 2 + 4(n + 1). 

.-.cos^i 

= i [- m ± Vm 2 + 4 (n + 1)]. 

66. Given sin A = m sin JB, and 

tan A = n tan 5 ; find sin .4 and 

cosl?. 

tan A = n tan JB. 

sin A sin S 

7 = % ^* 

cos^t cosJ5 

m sin jB _ n sin JB 
cos A cos J5 

cos A — — cos -B. 
n 

cos 2 A = — cos 2 B. 
n 2 

sin 2 A = m 2 sin 2 5. 

w 2 
1 = — cos 2 jB + m 2 sin 2 £. 

n 2 
w 2 
— cos 2 B + m 2 (1 - cos 2 J5) = 1. 



cos 2 B 



\-m 2 __ (1 - m 2 ) ?i 2 
~m 2 ~ (1 -7i 2 )m 2 ' 



COS B : 



m 



4 



■n? 



_ m 2 ^ 1 — m 2 
cos 2 ^4. = — cos 2 2? = 

7l 2 1 — ft 2 



sin 2 ^4 = 1 



1 — ra 2 _ m 2 — n 2 
1 — n 2 1 — n 2 



sin^4 



Vm 2 — ri 
1-n 2 



67. If tan 2 J. + 4 sin 2 J. = 6, 

find ^4. 

tan 2 ^4 + 4 sin 2 ^4 = 6. 

sin 2 .A , , . . A _ 
+ 4 sin 2 .4 = 6. 



1 -sin 2 J. 
sin 2 A + 4 sin 2 .4 — 4 sin 4 J. 

= 6 -6 sin 2 .A. 



4 sin 4 ^4-11 sin 2 ^4 + 6 = 0. 
(4 sin 2 J. - 3) (sin 2 .4 - 2)= 0. 

sin 2 ^4 = f or 2^ 

sin A = ± -J- V3. 
.\A=± 60°, ± 120°. 

68. If sin A = sin 2 -4, find ^4. 

sin J. = sin 2 .4 = 2 sin ^4 cos ^4. 
.-. sin A(l -2 cos .4) = 0. 
.*. sin A — 0, 
or 1 — 2 cos A — 0. 

.-. cos J. = £. 
^4 = 0°, 180°, ± 60°. 

69. If tan 2 ^4 = 3 tan A, find 4. 

tan 2 ^4 = 3 tan ^4. 

2 tan J. 
= 3 tan -4 . 

1 -tan 2 ^4 

2 tan A = 3 tan A - 3 tan 3 ^4. 

3 tan 3 J. -tan A = 0. 

tan ^4 (3 tan 2 A - 1) = 0. 

tan J. = 0, 

or 3 tan 2 .4-1 = 0. 

.-. tan.4 = ± 1 V3. 
-4 = 0°, 180°, 30°, 150°, 210°, 330°. 

70. Prove that tan 50° + cot 50° 
= 2 sec 10°. 

tan 50° + cot 50° = tan 50° + 

tan 50° 

_ tan 2 50° + 1 

~~ tan 50° 

sec 2 50° 



tan 50° 
1 



sin 50° cos 50° 

2 
2 sin 50° cos 50° 
2 



sin 100° 

2 
cos 10° 
2 sec 10°, 



TEACHEKS' EDITION. 



145 



71. Given a regular polygon of 
n sides, and calling one of them a, 
find expressions for the radii of the 
inscribed and the circumscribed cir- 
cles in terms of n and a. 

If P, H, D are the sides of a regu- 
lar inscribed pentagon, hexagon, 
and decagon, prove P 2 = H 2 + D 2 . 

(i) Angle subtended by each side 

•360° 

a at the centre of the circle is 

n 

Hence, if R is the radius of the 
circumscribed circle, and r that of 
the inscribed circle, 



ia 

R ' 
la 

r 



tan 



.-. R = - esc 

2 



180° 

L J 

180° 
n 

180° 



-cot 



n 
180° 



(ii) Let R 

P 

H 

B 

To prove P 2 

or 4 sin 2 36° 

Now sin 36° 

or sin (2 x 18°) 

By [12] and 

2 sin 18° cos 18° 
2 sin 18° 



.-. 4 sin 2 18° 
l + 4sin 2 18° 
By [16], 



= 1 ; then 
= 2 sin 36°. 
= 2 sin 30° = 1. 

= 2 sin 18°. 

= # 2 + D 2 , 

= 1 + 4 sin 2 18°. 

= cos 54°, 

= cos (3x18°). 

Prob. 19, Ex. XIV, 

=4cos 3 18°-3cosl8°. 
= 4 cos 2 18° - 3 
= 4 -4sin 2 18°-3 
= l-4sin 2 18°. 
= 1 - 2 sin 18° 
= 1—2 cos 72°. 
= 2 - 2 cos 72° 
= 2 (1 - cos 72°) 
= 4 sin 2 36°. 



72. Obtain the formula for the 
area of a triangle, given two sides 
6, c and the included angle A. 

Let p be the length of the perpen- 
dicular from B on b. Then 

F = ipb. 
But p = csin-4. 

.-. F = i c sin A x b 
= ibc sin^.. 

73. Obtain the formula for the 
area of a triangle, given two angles 
A, B, and the included side c. 

sin^L 

a = c 

sin C 

sin P 





b = c^-~ • 
sin C 




p = 1 a& sin O 




, _ sin A sin B 

= -J-c 2 ' 

sin C 




i ^ sin JL sin B 
sin(^4+P) 


74. Obtain the formula for the 


area of a 

sides. 


triangle, given the three 


F = 


i ac sin B. 


By [12] 
sin B = 


2 sin -j- B cos J P. 


By [28] 


? 


sin %B = 


j(s -a)(s- c) 

V ac 


By [29] 


? 



COS -J- P : 



f s (s - 6) 



.-. sinP = — Vs (s — a)(s — b)(s — c). 
.-. P = Vs (s — a)(s — b) (s — c). 

75. If a is the side of an equilateral 

, . . a 2 V3 

triangle, show that its area is — - — . 



146 



PLANE TRIGONOMETRY. 



F = i be sin A 
= i a 2 sin 60° 

= ia 2 xiV3 
a 2 V3 



76. Two consecutive sides of a 
rectangle are 52.25 chains and 38.24 
chains. Find the area. 

Area = 52.25 x 38.24. 

log 52.25 = 1.71809 

log 38.24 = 1.58252 

log area = 3.30061 

Area = 1998. 

1998 sq. ch. = 199 A. 8 sq. ch. 

77. Two sides of a parallelogram 
are 59.8 chains and 37.05 chains, 
and the included angle is 72° 10'. 
Find the area. 

Area = 59.8 x 37.05 sin 72° 10'. 

log 59.8 = 1.77670 

log 37.05 = 1.56879 

log sin 72° 10 / = 9.97861 

log area = 3.32410 

Area = 2109.1. 
2109.1 sq. ch. = 210 A. 9.1 sq. ch. 

78. Two sides of a parallelogram 
are 15.36 chains and 11.46 chains, 
and the included angle is 47° 30'. 
Find the area. 

Area = 15.36 x 11.46 sin 47° 30'. 

log 15.36 = 1.18639 

log 11.46 = 1.05918 

log sin 47° 30' = ^86763 

log area = 2.11320 

Area = 129.78. 

129.78 sq. ch. = 12 A. 9.78 sq. ch. 



79. Two sides of a triangle are 
12.38 chains and 6.78 chains, and 
the included angle is 46° 24'. Find 
the area. 

Area = i x 12. 38 x 6. 78 sin 46° 24'. 

log 6. 19 = 0.79169 

log 6. 78 = 0.83123 

log sin 46° 24' = 9.85984 

log area = 1.48276 

Area = 30.392. 
30.392 sq. ch. = 3 A. 0.392 sq. ch. 

80. Two sides of a triangle are 
18.37 chains and 13.44 chains, and 
they form a right angle. Find the 
area. 

Area = ix 18.37x13.44. 

colog2 = 9.69897 -10 
log 18.37 = 1.26411 
log 13.44 = 1.12840 
log area = 2.39251 

Area = 123.45. 
123.45 sq. ch. = 12 A. 3.45 sq. ch. 

81. Two angles of a triangle are 
76° 54' and 57° 33' 12", and the in- 
cluded side is 9 chains. Find the 
area. 

From [34], 

_ 9 2 sin 76° 54' sin 57° 33'12" 
r6a ~ 2 sin 134° 27' 12" 

log 81 = 1.90849 

log sin 76° 54' = 9.98855 

log sin 57° 33' 12" = 9.92629 

colog2 = 9.69897-10 
colog sin 134° 27' 12" = 0.14641 
log area = 1.66871 

Area = 46.634. 
46.634 sq. ch. = 4 A. 6.634 sq. ch. 



teachers' edition. 



147 



82. Two sides of a triangle are 
19.74 chains and 17.34 chains. The 
first bears N. 82° 30' W. ; the second, 
S. 24° 15' E. Find the area. 

Included angle = 121° 45'. 
Area = \ x 19.74 x 17.34 sin 121° 45'. 
colog 2 = 9.69897 - 10 
log 19.74 = 1.29535 
log 17.34 = 1.23905 
log sin 121° 45' = 9.92960 
log area = 2.16297 
Area = 145.54. 
145.54 sq. ch. = 14 A. 5.54 sq. ch. 

83. The three sides of a triangle 
are 49 chains, 50.25 chains, and 
25.69 chains. Find the area. 



Area = Vs (s — a)(s — b) (s — c). 
s = i (49 + 50.25 + 25.69) 
= 62.47. 

s-a = 13.47. 
s-b = 12.22. 
s-c =36.78. 

log 62.47 = 1.79567 

log 13.47 = 1.12937 

log 12.22 = 1.08707 

log 36. 78 = 1.56561 

2 )5.57772 

log area = 2.78886 

Area = 614.97. 

614.97 sq. ch. = 61 A. 4.97 sq. ch. 

84. The three sides of a triangle 
are 10.64 chains, 12.28 chains, and 
9 chains. Find the area. 
s = £(10.64 + 12.28 + 9) 
= 15.96. 

s-a = 5.32. 
s-b = 3.68. 
s-c = 6.96. 



log 15.96 = 1.20303 

log 5.32 =0.72591 

log 3.68 = 0.56585 

log 6.96 = 0.84261 

2 )3.33740 

log area = 1.66870 

Area = 46.633. 

46.633 sq. ch. = 4 A. 6.633 sq. ch. 

85. The sides of a triangular field, 

of which the area is 14 acres, are in 

the ratio of 3, 5, 7. Find the sides. 

Let the sides, measured in chains, 

be 3x, 5x, 7x. 

14 A. = 140 sq. ch. 
Then s = £(3x + 5x + 7x) 

= 7.5*. 
s — a = 4.5 x. 
s - b = 2.5x. 
s — c = 0.5 x. 



= V7.0X X 
~-2 


4.5x x 2.5x 


x 0.5x 


= — Vl5 x 9 x 5 
4 




= ^V3. 
4 

.-.x 2 


_ 4 x 140 _ 
15 V3 


112 
3V3 


log 112 

colog 3 V3 

< 


= 2.04922 

= 9.28432 - 
2)1.33354 


10 


logx 


= 0.66677 





x = 4.6427. 

3x = 13.9281. 

5x = 23.2135. 

7x = 32.4989. 

Sides are 13.93 chains, 23.21 

chains, 32.50 chains. 

86. In the quadrilateral ABCD 
we have AB, YJ.22 chains; AD, 



148 



PLANE TRIGONOMETRY. 



7.45 chains; CD, 14.10 chains ; BC, 
5.25 chains ; and the diagonal A C, 
15.04 chains. Find the area. 
In the triangle ABC, 
s = -£-(17.22 + 5.25 + 15.04) 
= 18.755. 

s-a= 1.535. 
s-b = 13.505. 
s-c = 3.715. 

log 18.755 = 1.27312 
log 1.636 = 0.18611 
log 13.505 = 1.13049 
log 3.715 = 0.56996 
2 )3.15968 
log area = 1.57984 

Area = 38.005. 
In the triangle A CD, 

s = i(15.04 + 14.10 + 7.45) 
= 18.295. 

s-a= 3.255. 
s-6 = 4.195. 
s-c = 10.845. 

log 18.295 = 1.26233 
log 3.255 = 0.51255 
log 4.195 = 0.62273 
log 10.845 = 1.03523 
2 )3.43284 
log area = 1.71642 
Area = 52.050. 
Area ABC =38.005 
Area .4 CD = 52.050 
Area ABCD = 90.055 
90.055 sq. ch. = 9 A. 0.055 sq. ch. 

87. The diagonals of a quadri- 
lateral are a and b, and they inter- 
sect at an angle D. Show that the 
area of the quadrilateral is i ab sin D. 

Let the parts into which the diago- 
nals are divided by their intersec- 



that 



tion be a\, a 2f and &i, b 2 , so that 
a = ai + a 2 and b = 6 X + b 2 . Then 
the areas of the four triangles into 
which the diagonals divide the quad- 
rilateral are 

i ai&i sin D, i a 2 bi sin D, 
■J ciib 2 sin D, $ a 2 b 2 sin D. 

The area of the quadrilateral is 
therefore 

i «i (h + h) sin D + £ a 2 (&i + 6 2 ) sin D 
= i (ai + a 2 ) (&i + 6 2 ) sin D 
= i ab sin D. 

88. The diagonals of a quadri- 
lateral are 34 and 56, intersecting 
at an angle of 67°. Find the area. 
Area = J- x 34 x 56 x sin 67°. 
log 17 = 1.23045 
log 56 = 1.74819 
log sin 67° = 9.96403 
log area = 



2.94267 
Area = 876.34. 






89. The diagonals of a quadri- 
lateral are 75 and 49, intersecting 
at an angle of 42°. Find the area. 
colog2 = 9.69897 -10 
log 75 ='1,87606 
log 49 = 1.69020 
log sin 42° = 9.82551 
log area = 3.08974 
Area = 1229.5. 



90. Show that the area of a regu- 
lar polygon of n sides, of which one 

is a, is 



nd 



2 x 180° 

cot 

4 n 



Lines joining the vertices to the 
centre divide the polygon into n 
equal isosceles triangles, the bases 
of which are a, and the vertical 



TEACHEKS EDITION. 



149 



360° 

angles The altitude of each 

° n 

triangle is 



h 



a 180° 

: - COt 

2 n 



and the area of each is 



£ ah = — cot 
4 



180° 



Hence, the area of the polygon is 

na 2 A 180° 

— cot 

4 n 

91. One side of a regular penta- 
gon is 25. Find the area. 

5 x 25 2 A 180° 
Area = cot 

4 5 

= 781.25 cot 36°. 

log 781.25= 2.89279 

log cot 36° = 10.13874 

log area = 3.03153 

Area = 1075.3. 

92 . One side of a regular hexagon 
is 32. Find the area. 

6 x 322 180 o 
Area = cot 

4 6 

= 1536 cot 30°. 
log 1536= 3.18639 
log cot 30° = 10.23856 
log area = 3.42495 
Area = 2660.4. 

93. One side of a regular decagon 
is 46. Find the area. 



Area . 



10 x 46 2 180° 

cot 

4 10 



= 5290 cot 18°. 
log 5290= 3.72346 
log cot 18° = 10.48822 
log area = 4.21168 
Area = 16,281. 



94. Find the area of a circle 
whose circumference is 74 feet. 

2 nr = 74. 
_37 

it 

o 37 2 

Area = nr 2 = 

it 

log37 2 = 3.13640 

colog it = 9.50285 - 10 

log area = 2.63925 

Area = 435.76. 

Area = 435.76 sq. ft. 

95. Find the area of a circle 
whose radius is 125 feet. 

Area = it x 125 2 . 
log 125 2 = 4.19382 

log it = 0.49715 
log area = 4.69097 

Area = 49,088. 

Area = 49,088 sq. ft. 



96. 


In a circle 


with a diameter 


of 125 feet find the 


area of a 


sector 


with an arc of 22°. 






Area of sector 


: area of 


circle 


= 22 


360. 






.-. area of sector = 


2 2 «r 1 1 2 5 
3 6 /t \ 2 


) 2 




= 


11 x 125 2 
720 


7t. 




log 11 = 


1.04139 






log 125 2 = 


4.19382 






colog 720 = 


7.14267 - 


10 




log it = 


0.49715 






log area = 


2.87503 






Area = 


749.95. 






Area = 


749.95 sq. 


ft. 



97. In a circle with a radius of 
44 feet find the area of sector with 
an arc of 25°. 



150 



Area = 



PLANE TRIGONOMETRY 



••Mr*.4# 



1210 7t 
9 



log 1210 = 3.08279 
log it = 0.49715 
colog 9 = 9.04576 - 10 
log area = 2.62570 

Area = 422.38. 
Area = 422.38 sq. ft. 

98. In a circle with a diameter of 
50 feet find the area of a segment 
with an arc of 280°. 

Area of segment = area of sector 
with same arc -f area of triangle 
with two sides equal to radius, and 
included angle of 80°. 

Area of sector = §f $ it 25 2 
4375 it 



log 4375 = 3.64098 
log7t = 0.49715 
colog 9 = 9.04576 - 10 
log area = 3.18389 

Area of sector 
Area of triangle 



1527.2. 

i x 25 2 sin 80° 

312.5 sin 80°. 



log 312.5 = 2.49485 

log sin 80° = 9.99335 

log area = 2.48820 

Area of triangle 
Area of segment 



307.75. 
1834.95 sq. ft. 



99. Find the area of a segment 
(less than a semicircle) of which the 
chord is 20, and the distance of the 
chord from the middle point of 
the smaller arc is 2. 




tan AEB = - 1 / = 5. 
log tan AEB = 10.69897. 

AEB = 78° 41' 24". 
ACB = l$0°- 2 AEB 
= 22° 37 / 12". 
AC = AB esc ACB 
= 10 esc 22° 37' 12" 



log 10 

log csc 22° 37' 12" 

log AC 

AC 
ACB 

Area of sector 



: 1.00000 
: 0.41497 
: 1.41497 




7T26 2 



log 377 

\0g1t 

log 26 2 
colog 3000 

log area = 2.42631 



360 

— _aUL Tt 26 2 

— 3 "' * u * 

= 2.57634 
= 0.49715 
= 2.82994 
= 6.52288 - 10 



Area of sector ■. 
Area of triangle 



= 266.88. 
CAB 
= ABxCB 
= 10 (26 - 2) 
= 240. 
Area of segment = 26.88. 



TEACHERS EDITION. 



151 



100. If r is the radius of a circle, 

the area of a regular circumscribed 

180° 

polygon of n sides is nr 2 tan 

n 

The area of a regular inscribed 

. . n 2 . 360° 

polygon is - r 2 sin 

2 n 

Lines drawn from the vertices to 

the centre divide the polygon into 

n equal isosceles triangles, the bases 

of which are the sides of the poly- 

360° 
gon and the vertical angles 

In the circumscribed polygon, 

180° 

each side = 2r tan , and the 

n 

altitude of each triangle is r. 
Hence, the area of each triangle is 

180° 

r 2 tan , and the area of the 

n 

180° 

polygon nr 2 tan 

n 

In the inscribed polygon, each 

180° 

side = 2 r sin , and the altitude 

n 

180° 

of each triangle is r cos Hence, 

n 

the area of each triangle is 

. . 180° 180° r 2 . 360° 

r- sm cos = — sm , 

n n 2 n 

and the area of the polygon is 

nr 2 . 360° 

I — sin 

, 2 n 

101. If a is a side of a regular 
polygon of n sides, the area of the 

inscribed circle is — cot 2 — . 
4 n 

The area of the circumscribed 

circle is ^csc*^!. 
4 n 

If r is the radius of the inscribed 

! circle, 180° 

a = 2 r tan 

n 



a 4 180° 

,\ r = - cot 

2 n 

nr 2 — cot 2 

4 n 

If E is the radius of the circum- 
scribed circle, 

oz> • 180° 
a = 2 R sin 



„ a 180° 
.-. R = - esc 

2 n 

7TR 2 = CSC 2 

4 ?i 



102. The area of a regular poly- 
gon inscribed in a circle is to that 
of the circumscribed regular polygon 
of the same number of sides as 3 to 4. 
Find the number of sides. 

n a . 3G0° . 180° _ . 

- r 2 sm : nr 2 tan = 3:4. 

2 n n 

o a • 3G0 ° o o + 180° 

2 nr 2 sin = 3 nr 2 tan • 

n n 



2 sin 



300° 



: 3 tan 



sm • 



180° 

n 
180° 

n 



. . 180° 180° 

4 sm cos = 3 ■ 

n n 180° 



cos- 

n 

cos = W3. 



180° 



= 30°. 



n = 6. 

103. The area of a regular poly- 
gon inscribed in a circle is the geo- 
metric mean between the areas of 
an inscribed and a circumscribed 
regular polygon of half the number 
of sides. 



152 



PLANE TRIGONOMETRY. 



Area of inscribed polygon of 2?i 



sides 



180° 



Area of inscribed polygon of n 

sides n . . 360° 

= - r 2 sin . 

2 n 

Area of circumscribed polygon of 



nr 2 tan 



180° 
n 

180° 



n . . 360° 

- r 2 sin x nr 2 tan — 

2 n n 

n 2 r 4 



rVr* . 360° 180° 
= sm tan 

2 U n . 180° 

A . 180° 180° Sm w 

= ?i 2 r 4 sin cos 

n n 180° 



cos- 



: n¥ sin 2 



180° 

n 
180° \ 2 



n 



= ( nr 2 sm j 



104. The area of a circumscribed 
regular polygon is the harmonic 
mean between the areas of an in- 
scribed regular polygon of the same 
number of sides and of a circum- 
scribed regular polygon of half that 
number. 

Area of circumscribed polygon of 

2 n sides 0x 90° 

= a = 2 nr 2 tan 

n 

Area of inscribed polygon of 2 n 



sides 



180° 



Area of circumscribed polygon of 



n sides 



To prove 



nr 2 tan 



180° 



1 1 

6 + c = 



nr 2 sin 



1 + cos 



180 ( 
n 

180° 



+ • 



nr 2 tan 



180° 



180° 



O .2 90 ° 

2 cos 2 — 



o 9 ' 90 ° 90 ° 

2 nr 2 sm — cos — 
n n 



o 90 ° 

2 cos — 

n 



2 nr 2 sin 



90° 



?=! + *. 

a 6 c 



« n 90° 

2 nr 2 tan — 
n 

__2 

~~ a' 

105. The perimeter of a circum- 
scribed regular triangle is double 
that of the inscribed regular triangle. 

Each side of circumscribed tri- 
angle = 2 r tan 60° = 2 V3 r. 
Each side of inscribed triangle 
= 2rsin60° = V3r. 

106. The square described about 
a circle is four-thirds the inscribed 
regular dodecagon. 

Area of square = 4 r 2 . 

*a a 12 9 • 360 ° 
Area of dodecagon = — r 2 sin 

= 6r 2 sin 30° = 3r 2 . 

107. Two sides of a triangle are 
3 and 12, and the included angle is 
30°. Find the hypotenuse of an 
isosceles right triangle of equal area. 



TEACHERS' EDITION. 



153 



Area of given triangle 

= i x 3 x 12 sin 30° = 9. 

Side of required triangle 
= V2 x 9 = 3 V2. 

Hypotenuse of required triangle 
= V2 (3 V2) 2 = V36 = 6. 

Required hypotenuse, 6. 

108. Taking the earth's equato- 
rial diameter to be 7925.6 miles, 
find the length in feet of the arc of 
one minute of a great circle. 
Circumference of great circle 

= it x 7925.6. 
Length of arc of 1/, in feet 
_ it x 7925.6 x 5280 
360 x 60 
7925.6 x 5280 it 



21600 

log 7925.6 = 3.89903 

log 5280 = 3.72263 

log 7t = 0.49715 

colog 21600 = 5.66555 - 10 

3.78436 

Arc of 1', 6086.4 feet. 

109= A ship sails from latitude 
43° 45' S., on a course N. by E. 2345 
miles. Find the latitude reached, 
and the departure made. 

Course = 11° 15'. 
Diff. lat. = 2345 cos 11° 15'. 
Depart. = 2345 sin 11° 15'. 

log 2345 = 3.37014 

log cos 11° 15' = 9.99157 

log. diff. lat. = 3.36171 

Diff. lat. = 2299.9' 

= 38° 19' 54". 



log 2345 = 3.37014 

log sin 11° 15' = 9.29024 

log depart. = 2.66038 

Depart. = 457.49. 
Latitude reached, 5° 25' 6" S.; 
departure, 457.49 miles. 

110. A ship sails from latitude 
1° 45' N., on a course S.E. by E., 
and reaches latitude 2° 3r S. Find 
the distance, and the departure. 

Course = 56° 15. 
Diff. lat. = 4° 16' = 256 miles. 

Dist. = 256 sec 56° 15'. 
Depart. = 256 tan 56° 15'. 
log 256 = 2.40824 
log sec 56° 15' = 0.25526 
log dist. =2.66350 
Dist. =460.79. 
log 256= 2.40824 
log tan 56° 15' = 10.17511 
log depart. = 2.58335 
Depart. =383.13. 
Distance, 460. 79 miles ; departure, 
383.13 miles. 

111. A ship sails from latitude 
13° 17' S., on a course N.E. by E. 
f E., until the departure is 207 
miles. Find the distance, and the 
latitude reached. 

Course = 64° 4r 15". 
Depart. = 207 miles. 
Dist. = 207 esc 64° 41' 15". 
Diff. lat. = 207 cot 64° 4V 15". 
log 207 = 2.31597 
log esc 64° 41 / 15" = 0.04384 
log dist. =2.35981 
Dist. = 228.98. 



154 



PLANE TRIGONOMETRY. 



cos course = 



log 207 = 2.31597 

log cot 64° 41' 15" = 9.67483 

logdiff. lat. = 1.99080 

Diff. lat. = 97.904' 

= l°37 / 54 ,/ . 
13° 17' - 1° 37' 54" = 11° 39' 6". 

Distance, 228.98 miles; latitude 
reached, 11° 39' 6" S. 

112. A ship, sails on a course 
between S. and E. 244 miles, leav- 
ing latitude 2° 52 / S. , and reaching 
latitude 5° 8' S. Find the course 
and the departure. 

Diff. lat. = 2° 16' = 136 miles. 
Dist. = 244 miles. 
136 
244* 

Depart. = V244 2 - 136' 2 
= V (244 + 136) (244 - 136) 
= V380x 108. 

log 136 = 2.13354 
colog 244 = 7.61261 - 10 
log cos course = 9.74615 

Course = 56° T 32". 

log 380 = 2.57978 
log 108 = 2.03342 
2 )4.61320 
log depart. = 2.30660 
Depart. =202.58. 
Course, S. 56° 7' 32" E. ; depart- 
ure, 202.58 miles. 

113. A ship sails from latitude 
32° 18' N., on a course between N. 
and W., a distance of 344 miles, and 
a departure of 103 miles. Find the 
course, and the latitude reached. 

Dist. = 344 miles. 
Depart. =103 miles. 



sin course s 



103 
344* 



Diff. lat. = V344* - lb3 2 
= V(344 + 103) (344 - 103) 
= V447 x 241. 

log 103 = 2.01284 
colog 344 = 7.46344 - 10 
log sin course = 9.47628 

Course = 17° 25 / 22". 

log 447 = 2.65031 

log 241 = 2.38202 

2 )5.03233 

logdiff. lat. =2.51616 

Diff. lat. = 328.22' 

= 5° 28' 13". 
32° 18' + 5° 28' 13" = 37° 46' 13". 

Course, N. 17° 25' 22" W. ; lati- 
tude reached, 37° 46' 13" N. 

114. A ship sails on a course be- 
tween S. and E. , making a difference 
of latitude 136 miles, and a depart- 
ure 203 miles. Find the distance, 
and the course. 

Diff. lat. = 136 miles. 
Depart. = 203 miles. { 
203 
136* 
log 203= 2.30750 
colog 136 = 7.86646 - 10 
log tan course = 10.17396 

Course = 56° 10' 49". 
Dist. = 203 esc 56° 10' 49". 
log 203 = 2.30750 
log esc 56° 10' 49" = 0.08051 
log dist. =2.38801 
Dist. = 244.35. 
Course, S. 56° 10' 49" E. ; dis- 
tance, 244.35 miles. 






tan course : 






TEACHERS EDITION. 



151 



115. A ship sails due north 15 
statute miles an hour for one day. 
What is the distance in a straight 
line from the point left to the point 
reached ? (Take earth's radius. 
39G2.8 statute miles.) 
Distance sailed in one day 

= 24 x 15 miles = 360 miles 
360 



2tt x 3962.8 
64800° 



x 360° 



3962.8 it 

log 64800 = 4.81158 
colog 3962.8 = 6.40200 - 10 
colog it = 9.50285- 10 
logdist. =0.71643 
Distance sailed = 5.2051° 

= 5° 12' 18". 
Chord of arc sailed 

■ =2 x 3962.8sin2°36 / 9 // 
= 7925.6 sin 2° 36' 9". 
log 7925.6 = 3.89903 
log sin 2° 36' 9" = 8.65712 
log chord = 2.56615 
Chord = 359.87. 
Required distance, 359.87 miles. 

116. Given the departure be- 
tween any two meridians at any 
latitude ; find the difference of lon- 
gitude of any point on one meridian 
from any point on the other. 
P 




In rt. A ODA, A 4 02) = 90° -lat. 
Hence, 

DA 

OA 

The A DAB and OEQ are similar. 
Therefore, 



: sin (90° -lat.) = coslat. 



DA 



AB 



_AB DA _ 

OE ~ EQ ° r ~OA ~ ~EQ' 

AB 

Substituting, cos lat. = 

EQ 
Therefore, 

A 7? 

EQ = = ABx sec lat. 

cos lat. 

That is, 

Din\ long. = depart, x sec lat. 



117. A ship in latitude 42° 16' 
N., longitude 72° 16' W., sails due 
east a distance of 149 miles. What 
is the position of the point reached ? 

Diff. long. = 149 sec 42° 16'. 

log 149 = 2.17319 
log sec 42° W = 0.13076 
log. diff. long. = 2.30395 

Diff. long. = 201.35 / = 3°21 / 21 // . 
Longitude of position reached, 
68° 54' 39" W. 

118. A ship in latitude 44° 49 r S., 
longitude 119° 42 7 E., sails due west 
until it reaches longitude 117° 16' E. 
Find the distance made. 

Diff. long. = 2° 26' = 146 7 . 
Depart. = 146 cos 44° 49 7 . 

log 146 = 2.16435 

log cos 44° 49' = 9. 85087 

log depart. =2.01522 

Depart. = 103.57. 
Distance made, 103.57 miles. 



156 



PLANE TRIGONOMETRY. 



119. A ship leaves latitude 31° 14' 
N., longitude 42° 19' W., and sails 
E.N.E. 325 miles. Find the posi- 
tion reached. 

Course = 67° 30'. 
Diff. lat. = 325 cos 67° 30'. 

log 325 = 2.51188 

log cos 67° 30' = 9.58284 

log diff. lat. =2.09472 

Diff. lat. = 124. 37' = 2° 4' 22 // . 
Mid. lat. = 32° 16' 11". 
Depart. = 325 sin 67° 30'. 
Diff. long. 

= 325 sin 67° 30' sec 32° 16' 11". 

log 325 = 2.51188 

logsin67°30 / = 9.96562 

logsec32°16 / ll // = 0.07286 

log. diff. long. = 2.55036 

Diff. long. = 355. II 7 = 5° 55' 7". 

Latitude of position reached, 33° 

18' 22" N. ; longitude, 36° 23' 53" W. 

120. Find the bearing and dis- 
tance of Cape Cod from Havana. 
(Cape Cod, 42° 2' N., 70° 3' W. ; 
Havana, 23° 9' N., 82° 22' W.) 

Diff. long. = 12° 19' = 739'. 

Diff. lat. = 18° 53' = 1133'. 

Mid. lat. = 32° 35' 30". 

Depart. = diff. long, x cos mid. lat. 

= 739 cos 32° 35' 30". 

depart. 

tan course = — 

diff. lat. 

_ 739 cos 32° 35' 30" 
1133 

log 739 = 2.86864 
log cos 32° 35' 30" = 9.92559 

colog 1133 = 6.94577 - 10 
log tan course = 9. 74000 



Course = 28° 47' 26". 
Dist. = diff. lat. x sec course 
= 1133 sec 28° 47' 26". 

log 1133 = 3.05423 

log sec 28° 47' 26" = 0.05730 

log dist. =3.11153 

Dist. = 1292.8. 
Bearing, N. 28° 47' 26" E. ; dis- 
tance, 1292.8 miles. 

121. Leaving latitude 49° 57' N. , 
longitude 15° 16' W., a ship sails 
between S. and W. till the depart- 
ure is 194 miles, and the latitude is 
47° 18' N. Find the course, dis- 
tance, and longitude reached. 
Diff. lat. = 2° 39" = 159 miles. 
Mid. lat. = 48° 37' 30". 
Depart. = 194 miles. 
Diff. long. = 194 sec 48° 37' 30". 
log 194 = 2.28780 
log sec 48° 37' 30" = 0.17981 
log diff. long. = 2.46761 
Diff. long. = 293.50' 

= 4° 53' 30". 
194 
159* 

log 194= 2.28780 
colog 159 = 7.79860 - 10 
log tan course = 10.08640 

Course = 50° 39' 44". 
Dist. = 159 sec 50° 39' 44". 
log 159 = 2.20140 
log sec 50° 39' 44" = 0.19799 
log dist. = 2.39939 
Dist. = 250.84. 
Course, S. 50° 39' 44" W. ; dis- 
tance, 250.84 miles; longitude 
reached, 20° 9' 30" W. 






tan course = 



TEACHERS' EDITION. 



157 



122. Leaving latitude 42° 30' N. , 
longitude 58° 51' W., a ship sails 
S.E. by S. 300 miles. Find the 
position reached. 

Course = 33° 45'. 

Diff. lat. = 300 cos 33° 45'. 

log 300 = 2.47712 

log cos 33° 45' = 9.91985 

log diff. lat. =2.39097 

Diff. lat. = 249.44' 

= 4° 9' 26". 
Mid. lat. = 40° 25' 17" . 
Depart. = 300 sin 33° 45'. 
Diff. long. 

= 300 sin 33° 45' sec 40° 25' 17". 

log 300 = 2.47712 

log sin 33° 45' = 9.74474 

log sec 40° 25' 17" = 0.11845 

log diff. long. = 2.34031 

Diff. long. =218.93' 

= 3° 38' 50". 
Latitude of position reached, 38° 
20' 34" N. ; longitude, 55° 12' 4" W. 

123. Leaving latitude 49° 57' N. , 
longitude 30° W., a ship sails S. 
39° W., and reaches latitude 47° 44' 
N. Find the distance, and longitude 
reached. 

Course = 39°. 
Diff. lat. = 2° 13' = 133 miles. 
Mid. lat. = 48° 50' 30". 
Dist. = 133 sec 39°. 
log 133 = 2.12385 
log sec 39° = 0.10950 
log dist. =2.23335 
Dist. = 171.14. 
Depart. = 133 tan 39°. 
Diff. long. 

= 133 tan 39° sec 48° 50' 30". 



cos course = ■ 



log 133 = 2.12385 

log tan 39° = 9.90837 

log sec 48° 50' 30" = 0.181C8 

log diff. long. = 2.21390 

Diff. long. = 103.64' 

= 2° 43' 38". 

Distance. 171.14 miles; longitude 
reached, 32° 43' 38" W. 

124. Leaving latitude 37° X., 
longitude 32° 16' W\, a ship sails 
between N. and W. 300 miles, and 
reaches latitude 41° N. Find the 
course, and longitude reached. 

Diff. lat. = 4° = 240 miles. 
Mid. lat. = 39°. 
Dist. = 300. 

240 
300* 

log 240 = 2.38021 
colog 300 = 7.52288 - 10 
log cos course = 9.90309 

Course = 3 6° 52' 12". 
Depart. = V300 2 - 2 40* 
= V60 x 540 
= 180. 
Diff. long. = 180 sec 39°. 

log 180 = 2.25527 

log sec 39° = 0.10950 

log diff. long. =2.36477 

Diff. long. =231.62' 

= 3° 51' 37". 
Course, N. 36° 52' 12" W. ; longi- 
tude reached, 36° 7' 37" W. 

125. Leaving latitude 50° 10' S., 
longitude 30° E., a ship sails E.S.E., 
making a departure of 160 miles. 
Find the distance, and position 
reached. 



158 



PLANE TRIGONOMETRY. 



Course = 67° 30'. 
Depart. = 160 miles. 

Dist. = 160 esc 67° 30'. 
Diff. lat. = 160 cot 67° 30'. 

log 160 = 2.20412 

log esc 67° 30' = 0.03438 

log dist. =2.23850 

Dist. = 173.18. 

log 160 = 2.20412 

log cot 67° 30' = 9.61722 

log diff. lat. = 1.82134 

Diff. lat. = 66.273' 

= 1° 6' 16". 
Lat. reached = 51° 16' 16". 

Mid. lat. = 50° 43" 8". 
Diff. long. = 160 sec 50° 43' 8". 

log 160 = 2.20412 

log sec 50° 43' 8" = 0.19851 

log diff. long. =2.40263 

Diff. long. = 252. 71' 

= 4° 12' 43". 
Distance, 173.18 miles; latitude 
of position reached, 51° 16" 16" S. ; 
longitude, 34° 12' 43" E. 

126. Leaving latitude 49° 30' N. , 
longitude 25° W., a ship sails be- 
tween S. and E. 215 miles, making 
a departure of 167 miles. Find the 
course, and position reached. 

167 
215* 



sin course : 



log 167 =2.22272 
colog 215 = 7.66756 - 10 
log sin course = 9. 89028 

Course = 50° 57' 48". 
Diff. lat. = V 215 2 - 16 T 2 
= V48 x 382. 






log 48 = 1.68124 

log 382 = 2.58206 

2 )4.26330 

log diff. lat. =2.13165 

Diff. lat. = 135.41' 

= 2° 15' 25". 
Mid. lat. =48° 22' 18". 
Diff. long. = 167 sec 48° 22' 18". 

log 167 = 2.22272 

log sec 48° 22' 18" = 017764 

log diff. long. =2.40036 

Diff. long. = 251.39' 

= 4° 11' 23". 

Course, S. 50° 57'48"E.; latitude 
of position reached, 47° 14' 35" N. ; 
longitude, 20° 48' 37" W. 

127. Leaving latitude 43° S. , lon- 
gitude 21° W., a ship sails 273 miles, 
and reaches latitude 40° 17' S. What 
are the two courses and longitudes 
which will satisfy the data ? 

The two courses make equal angles 
with the meridian on opposite sides. 

Diff. lat. = 2° 43' = 163 miles. 
Dist. = 273 miles. 

163 
cos course = — . 
273 

log 163 = 2.21219 
colog 273 = 7.56384 - 10 
log cos course = 9.77603 

Course = 53° 20' 21". 
Depart. = V273 2 - 163"2 
= Vll0 x436. 
Mid. lat. = 41° 38' 30". 
Diff. long. 

= VllO x 436 sec 41° 38' 30". 



TEACHERS' EDITION. 



159 



log VllO = 1.02069 

log V436 = 1.31075 

log sec 41° 38' 30" = 0.12050 

logdiff. long. =2.46604 

Diff. long. = 293.05' 

= 4° 53' 3". 
(i) Course, N. 53° 20' 21" E.; 
longitude of position reached, 
16° 6' 57' W. 

(ii) Course, N. 53° 20' 21" W.; 
longitude of position reached, 
25° 53' 3" W. 

128. Leaving latitude 17° N., 
longitude 110° E., a ship sails 219 
miles, making a departure of 162 
miles. What four sets of answers 
do we get ? 

The four courses all make the 
same angle with the meridian. 

162 

sin course = 

210 

log 162 = 2.20952 
colog219 = 7.65956 



10 



log sin course = 9. 86008 

Course = 47° 42' 33". 
Diff. lat. = V219 2 - 102 2 



= V57 x 381. 

log 57 = 1.75587 

log 381 = 2.58092 

2 )4.33679 

logdiff. lat. =2.16840 

Diff. lat. = 147.37' 

= 2° 27' 22". 
(i) Mid. lat. = 18° 13' 41". 
Diff. long. = 162 sec 18° 13' 41". 
log 162 = 2.20952 
log sec 18° 13' 41" = 0.02236 
log diff . long. =2.23188 



Diff. long. = 170.56' 

= 2° 50' 34". 
(ii) Mid. lat. = 15° 46' 19". 

Diff. long. = 162 sec 15° 46' 19". 
log 162 = 2.20952 
log sec 15° 46' 19" = 0.01667 
logdiff. long. =2.22619 

Diff. long. = 168.34' 

= 2° 48' 20". 

(i) Course, N. 47° 42' 33" E. ; lati- 
tude of position reached, 19° 27' 22" 
N., longitude, 121° 50' 34" E. 

Course, N. 47° 42' 33" W. ; latitude 
of position reached, 19° 27' 22" N., 
longitude, 116° 9' 26" E. 

(ii) Course, S. 47° 42' 33" E. ; lati- 
tude of position reached, 14° 32' 38" 
N., longitude, 121° 48' 20" E. 

Course, S. 47° 42' 33" W. ; latitude 
of position reached, 14° 32' 38" N., 
longitude, 116° 11' 40" E. 

129. A ship in latitude 30° sails 
due east 360 statute miles. What 
is the shortest distance from the 
point left to the point reached ? 

Solve the same problem for lati- 
tudes 45°, 60°. 

By Prob. 116, Ex. XXIII, radius 
of parallel 

= 3962.8 cos lat. 
Arc sailed, in degrees, 

360 x 360 
~~2 7t x 3962. 8 cos lat.' 

Iog360 2 = 5.11260 
colog27T = 9.20182 - 10 
colog 3962.8 = 6.40200-10 
0.71642 

Arc sailed, in degrees, 

= 5.205 sec lat, 



160 



PLANE TRIGONOMETRY. 






Arc sailed, in minutes, 

= 312.3 sec lat. 
Chord of arc 

= 2 rad. of parallel sin (i arc) 
= 2 x 3962.8 cos lat. 

sin (156.15' sec lat.) 
= 7925.6 cos lat. 

sin (156.15' sec lat.). 

(i) lat. = 30°. 

log 156.15 = 2.19354 

log sec 30° = 0.06247 

log i arc = 2.25601 

iarc = 180.30' 

= 3° 0' 18". 

log 7925.6 = 3.89903 

log cos 30° = 9.93753 

logsin3°0'18" = 8.71952 

log chord = 2.55608 

Chord = 359.82. 

(ii) lat. = 45°. 

log 156.15 = 2.19354 

log sec 45° = 0.15051 

log i arc = 2.34405 

£arc = 220.82' 

= 3° 40' 49". 

log 7925.6 = 3.89903 

log cos 45° = 9.84949 

log sin 3° 40' 49" = 8.80746 

log chord = 2.55598 

Chord = 359.73. 

(iii) lat. = 60°. 

sec lat. = 2. 
£arc = 312.30' 

= 5° 12' 18". 

log 7925.6 = 3.89903 

log sin 5° 12' 18" = 8.95770 

log cos 60° = 9.69897 

log chord = 2.55570 

Chord = 359.50. 



Shortest distance, in lat. 30°, 
359.82 miles; in lat. 45°, 359.73 
miles; in lat. 60°, 359.50 miles; 
in general 7925.6 cos lat. x sin 
(156.15' sec lat.). 

130. Leaving latitude 37° 16' S., 
longitude 18° 42' W., a ship sails 
N.E. 104 miles, then N.N.W. 60 
miles, then W. by S. 216 miles. 
Find the position reached, and its 
bearing and distance from the point 
left. 

First course = 45°. 

Diff. lat. = 104 cos 45°. 
Depart. = 104 sin 45°. 

log 104 = 2.01703 

log cos 45° = 9.84949 

log diff. lat. = 1.86652 

Diff. lat. = 73.54 N. 
Depart. = 73.54 E. 
Second course = 22° 30'. 

Diff. lat. = 60 cos 22° 30'. 
Depart. = 60 sin 22° 30'. 

log 60 = 1.77815 

log. cos 22° 30' = 9.96562 

log diff. lat. =1.74377 

Diff. lat. = 55.434 N. 



log 60 : 

log sin 22° 30' : 

log depart. : 

Depart. : 

Third course : 

Diff. lat. : 

Depart. : 

log 216 : 

log cos 78° 45' : 

log diff. lat. : 

Diff. lat. : 



: 1.77815 
: 9.58284 
: 1.36099 

: 22.961 W. 
.- 78° 45'. 
: 216 cos 78° 45'. 
: 216 sin 78° 45'. 

: 2.33445 
: 9.29024 
: 1.62469 

: 42.14 S. 



teachers' edition. 



161 



log 216 = 2.33445 
log sin 78° 45' = 9.99157 
log depart. =2.32602 
Depart. = 211.85 W. 
Total diff. lat. = 86.834' N. 

= 1° 26' 50" N. 
Lat. reached = 35° 49' 10" S. 

Mid. lat. = 36° 32' 35". 
Total depart. = 161.271 W. 
Diff. long. = 161.271 sec 36° 32 , 35 // . 

log 161.271 = 2.20755 

log sec 36° 32' 35" = 0.09506 

log diff. long. =2.30261 

Diff. long. = 200.73 / 

= 3° 20' 44". 
Long, reached = 22° 2' 44" W. 

161.271 
tan course = 

86.834 

log 161.271= 2.20755 
colog 86.834 = 8.06131 - 10 
log tan course = 10.26886 
Course = 61° 42'. 
Dist. =86.834 sec 61° 42'. 
log 86.834 = 1.93869 
log sec 61° 42' = 0.32414 
log dist. =2.26283 
Dist. = 183.16. 
Course, N. 61° 42' W.; distance, 
183.16 miles; latitude reached, 35° 
49' 10" S.; longitude, 22° 2' 44" W. 

131. A ship leaves Cape Cod 

(Example 120), and sails S.E. by S. 

114 miles, N.byE. 94 miles, W.N.W. 

42 miles. Solve as in Example 130. 

First course = 33° 45'. 

Diff. lat. =114 cos 33° 45'. 
Depart. = 114 sin 33° 45'. 
log 114 = 2.05690 
log cos 33° 45' = 9.91985 
log diff. lat. = 1.97675 



Diff. lat. = 94.787 S. 

log 114 = 2.05690 

log sin 33° 45' = 9.74474 

log depart. = 1.80164 

Depart. = 63.334 E. 

Second course = 11° 15'. 

Diff. lat. = 94 cos 11° 15'. 
Depart. = 94 sin 11° 15'. 
log 94 = 1.97313 
log cos 11° 15' = 9.99157 
log diff. lat, = 1.96470 
Diff. lat. = 92.194 N. 
log 94 = 1.97313 
log sin 11° 15' = 9.29024 
log depart. = 1.26337 
Depart. = 18.339 E. 
Third course = 67° 30'. 

Diff. lat. = 42 cos 67° 30'. 
Depart. = 42 sin 67° 30'. 
log 42 = 1.62325 
log cos 67° 30' = 9.58284 
log diff. lat. = 1.20609 
Diff. lat. = 16.073 N. 

log 42 = 1.62325 
log sin 67° 30' = 9.96562 
log depart. = 1.58887 
Depart. = 38.804 W. 
Total diff. lat. = 13.48' N. 

= 13' 29" N. 
Lat. of C. Cod = 42° 2'. 
Lat. reached = 42° 15' 29" N. 

Mid. lat. = 42° 8' 44". 
Total depart. = 42.869 E. 

Diff. long. = 42.869sec42°8 / 44". 

log 42.869 = 1.63214 
log sec 42° 8' 44" = 0.12992 
log diff. long. = 1.76206 
Diff. long. = 57.817' E. 
= 57" 49" E. 



162 



PLANE TRIGONOMETRY. 



Long, of Cape Cod = 70° 3' W. 
Long, reached = 69° 5' 11" 

42.869 



W. 



tan course = ■ 



13.48 

1.63214 

8.87031 



10 



log 42. 869 = 
colog 13.48 = 
log tan course = 10.50245 

Course = 72° 32' 40"'. 
Dist. = 13.48 sec 72° 32' 40". 

log 13.48 = 1.12969 
log sec 72° 32' 40" = 0.52293 
log dist. =1.65262 
Dist. =44.939. 
Course N. 72° 32' 40" E. ; distance, 
44.939 miles ; latitude reached, 42° 
15' 29" N.; longitude, 69° 5' 11" W. 

132. A ship leaves Cape of Good 
Hope (latitude 34° 22' S., longitude 
18° 30' E.) and sails N.W. 126 
miles, N. by E. 84 miles, W.S.W. 
217 miles. Solve as in Example 130. 
First course = 45°. 

Diff. lat. = 126 cos 45°. 
Depart. = 126 sin 45°. 

log 126 = 2.10037 

log cos 45° = 9.84949 

log diff. lat. = 1.94986 

Diff. lat. = 89.096 N. 

Depart. = 89.096 W. 

Second course = 11° 15'. 

Diff. lat. = 84 cos 11° 15'. 
Depart. = 84 sin 11° 15". 

log 84 = 1.92428 

log cos 11° 15" = 9.99157 

log diff. lat. =1.91585 

Diff. lat. = 82.386 N. 

log 84 = 1.92428 

log sin 11° 15' = 9.29024 

log depart. = 1.21452 



Depart. : 

Third course : 

Diff. lat. : 

Depart. : 

log 217 : 

log cos 67° 30' : 

log diff. lat. : 

Diff. lat. : 

log 217: 
log sin 67° 30' = 
log depart. = 

Depart. = 

Total diff. lat. = 

Lat. reached = 

Mid. lat. = 

Total depart. = 

Diff. long. =273.198 

log 273. 198 = 

log sec 33° 37' 47" = 

log diff. long. = 

Diff. long. = 

Long, reached = 
tan course = 



: 16.388 E. 
: 67° 30'. 
: 217 cos 67° 30'. 
: 217 sin 67° 30'. 

: 2.33646 

: 9.58284 
: 1.91930 

: 83.042 S. 

: 2.33646 
: 9.96562 
: 2.30208 

: 200.49 W. 

: 88.440' N. 
: 1°28'26"N. 
: 32° 53' 34" S. 

33° 37' 47". 

273.198 W. 

sec 33° 37' 47". 

2.43648 
0.07954 
2.51602 

328.11' 
5° 28' 7". 
13° V 53" E. 
273.198 



3.44 



log 273. 198= 2.43648 
colog 88.44 = 8.05335 - 10 
log tan course = 10.48983 

Course = 72° 3' 43 
Dist. =88.44 sec 72° 3' 43' 

log 88.44 = 1.94066 

log sec 72° 3' 43" = 0.51147 
log dist. =2.45812 

Dist. = 287.16. 
Course, N. 72° 3' 43" W. ; distance, 
287.16 miles; latitude reached, 32° 
53' 34" S., longitude, 13° V 53" E. 






teachers' editiox. 163 



Exercise XXIY. Page 107. 

1. Prove that sin x 4- cos x = V2 cos (x — i it). 

By [9], cos (x — i it) — cos x cos \ it + sin x sin i it 

= \ V2 cos x + i "^2 sin x 
= \ V2 (cos x + sin x). 
.-. sin x + cos x = V2 cos (x — i it). 

2. Prove that sin x — cos x = — V2 cos (x + J it). 

By [5], cos (x + \ it) = cosx cos \ it — sin x sin J at 

= i V2 cos x — \ V2 sin x 
= \ V2 (cos x — sin x). 
.*. sin x — cos x — — V2 cos (x + i 7t). 

3. Prove that sin x -f V3 cos x = 2 sin (x 4- i 7t). 

By [4] , sin (x + J it) — sin x cos \ it 4- cos x sin i 7T 

= J sin x + 1 V3 cos x 
= \ (sin x + V3 cos x). 
.-. sin x + V3 cos x = 2 sin (x + i it). 

4. Prove that sin (x 4- i n) 4- sin (x — \ it) = sin x. 

By [4] , sin (x 4- % it) = sin x cos J it + cos x sin -J 7T 

= i sin x 4- i V.3 cos x. 
By [8], sin (x — \ it) — -$- sin x — J V3 cos x. 

.-. sin (x 4- i it) 4- sin (x — i it) = sin x. 

5. Prove that cos (x 4- J ?r) 4- cos (x — £ it) — V3 cos x. 
By [5], cos(x 4- \7t) = \ V3cosx — -J-sinx. 
By [9], cos (x — \ it) = i V3 cos x 4- i sin x. 

.-. cos (x 4- i ft) 4- cos (x — i 7T) = V3 cos x. 

6. Prove that tan x + sec x = tan {^x -\-\it). 

sin x 1 

tan x 4- sec x = f- 



cos x cos x 
sin x 4- 1 



By [16] and [12], 



cosx 
X — cos (x 4- j 7t) 

sin (x 4- i it) 

2sinH(ft + J7r) 
2 sin \ (x 4- i #) cos \ (x 4- -J it) 



164 PLANE TRIGONOMETRY. 

_ sin (jx + J7t) 

~ COS(iX^-iTT) 

= tan(|x + lit). 
7. Prove that tan x 4- sec x = ■ 



sec x — tan x 

By Prob. 2, Ex. V, sec 2 x = 1 + tan 2 x. 

sec 2 x — tan 2 x = 1. 

(sec x + tan x) (sec x — tan x) — 1. 

1 
.;. sec x -f tan x == 



8. Prove that 



sec x — tan x 
1 — tan x cot x — 1 



1 -f tan x cot x + 1 

1 



cot x — 1 tan x 1 — tan x 



cot x + 1 1 1 + tan x 

tanx 



^ ^ ^ sin x 1 + cos x 

9. Prove that 1 = 2 esc x. 

1 + cos x sin x 

sin x 1 + cos x _ sin 2 x + (1 + cos x) 2 



1 + cos x sin x sin x (1 + cos x) 

_ sin 2 x + cos 2 x + 2 cos x + 1 
sin x (1 + cos x) 

1 + 2 cos x + 1 



sinx(l + cosx) 
2 (1 -+ cos x) 





sin x (1 + cos x) 




2 






smx 






= 2 esc x. 




x + cot X = 


2 esc 2 x. 




tan x + cot 


sin x , cos x 
x = + - 





cos x sin x 
sin 2 x + cos 2 x 
sin x cos x 
1 



sin x cos x 



teachers' edition. 165 

2 



2 sin x cos x 

= 2 csc 2 x. 

11. Prove that cot x — tan x = 2 cot 2 x. 

cosx sin x cos 2 x — sin 2 x 2 cos 2 x 

cot x — tan x = = = — = 2 cot 2 x. 

sin x cosx sin x cos x sin2x 

12. Prove that 1 + tan x tan 2 x = sec 2 x. 

sin x sin 2 x 



1 + tan x tan 2 x = 1 + 

= 1 + 



cos x cos 2 x 
2 sin 2 x cosx 



cosx(l — 2 sin 2 x) 
_ 2 sin 2 x 

1 — 2sin 2 x 

1 



1 - 2sin 2 x 
1 



cos 2x 

== sec 2 x. 



sec 2 x 
13. Prove that sec 2 x = 



2 — sec 2 x 



1 1 cos 2 x sec 2 x 

sec 2 x = 



cos2x 2cos 2 x — 1 1 2 — sec 2 x 

cos 2 x 



14. Prove that 2 sec 2 x = sec (x -f 45°) sec (x — 45°). 

9 



2 sec 2 x = ■ 



cos 2x 
2 



cos 2 x — sin 2 x 
2 



By Probs. 1 and 2, Ex. XXIV, = 



(cos x — sin x) (cos x + sin x) 

2 



2 cos (x + 45°) cos (x — 45°) 
= sec (x + 45°) sec (x — 45°). 



166 PLANE TRIGONOMETRY. 

cos x + sin x 



15. Prove that tan 2 x -f sec 2x = ■ 



cos x-sinx 



x o sin 2x 1 

tan 2 x + sec 2 x = h 



By [16] and [12], 



By [4] and [5], 



cos 2 x cos 2 x 
sin 2 x + 1 

cos2x 
1 - cos (2 x + 90°) 

sin (2 x + 90°) 
2 sin 2 (x + 45°) 
" 2 sin (x + 45°) cos (x + 45°) 
sin (x + 45°) 
cos (x -f 45°) 

sin x cos 45° -f cos x sin 45° 
cos x cos 45° — sin x sin 45° 
\ V2 sin x + \ V2 cos x 

1 V2 cos x — i V2 sin x 
cos x + sin x 



16. Prove that sin 2 x : 



cos x — sin x 
2 tan x 



1 + tan 2 x 
2 tan x 2 tan x 



1 + tan 2 x sec 2 x 

2 sin 3 x 



= 2tanxcos 2 x = 2sinxcosx=sin2x. 



17. Prove that 2 sin x + sin 2 x = - 

1 — cos x 

2 sin x + sin 2 x = 2 sin x + 2 sin x cos x 

= 2 sinx(l + cosx). 

But 1 — cos 2 x = sin 2 x. 

sin 2 x 
.-. 1 + cos x = 



.\ 2 sin x + sin 2 x = 2 sin x 



1 — cos x 

sin 2 x 



1 — cos x 
2sin 3 x 



18. Prove that sin 3 x = 



1 — cos x 

sin 2 2x — sin 2 x 
sin x 



By [20] , sin 2 x -f sin x = 2 sin f x cos \ x. 

By [21] , sin 2 x — sin x = 2 cos § x sin J x. 



teachers' edition. 167 

.\ sin 2 2 x — sin 2 x = 2 sin § x cos § x x 2 sin i x cos J x 

By [12], = sin3xsinx. 

. _ sin 2 2 x — sin 2 x 
.*. sm 3 x = 



19. Prove that tan 3 x = 



smx 
3 tan x — tan 3 x 



1-3 tan 2 x 
tan 3 x = tan (2 x + x) 

_ tan 2 x + tan x 

1 — tan 2 x tan x 

2 tanx 



„ Mn 1 — tan 2 x 

By [14], 



4- tanx 



2 tanx 

tanx 



20. Prove that 
By [24], 



1 — tan 2 x 
_ 3 tan x — tan 3 x 
l-3tan 2 x 

tan 2 x + tan x _ sin 3 x 
tan 2 x — tan x sin x 

sin ^1 -f sin B _ tan |(^1 4- B) 



sin J. — sin B tan £ (J. — B) 

, ,. . . sin J. tan£M + J5) + tan-J-M - E) 

By composition and division, = — ■ 

sin B tan i(A + B)- tan i(A-B) 

_ . , . , ^ sin 3 x tan 2 x + tan x 

Put 3x for A, x for B, = 

sin x tan 2 x — tan x 

21. Prove that sin (x + y) + cos (x — y) = 2 sin (x -f J ar) sin (y -f £ tt). 
By [4], sin (x -f y) = sin x cos y + cos x sin ?/. 

By [9], cos (x — y) = cos x cos y + sin x sin y. 

.-. sin (x + y) + cos (x — y) = (sin x + cos x) cos y + (cos x -f- sin x) sin y 

= (sin x + cos x) (sin y + cos y). 
But sin x -f cos x = V2 (-J- V2 sin x -f -J V2 cos x) 

= V2sm(x + i7T). 
Similarly, sin y + cos y = V2 sin (y + J tt). 
.-. sin (x + 2/) + cos (x — y) = 2 sin (x + i #) sin (y + i it). 

22. Prove that sin (x -f y) — cos (x — y) = — 2 sin (x — I it) sin (y — I it). 
By [4], sin (x + y) = sin x cos y + cos x sin y. 

By [9], cos (x — y) = cos x cos ?/ + sin x sin ?/. 

.-. sin (x + y) — cos (x — y) = (sin x — cos x) cos y + (cos x — sin x) sin y 
— (sin x — cos x) (cos y — sin y) 
= — 2 sin (x — i it) sin (?/ — \ it). 



168 PLANE TRIGONOMETRY. 

sin (x -\- if\ 
23. Prove that tan x + tan y = — v y/ . 

cos x cos y 

sin x , sin 2/ 

tan x -f tan ?/ = — — -\ 

cos x cos y 

_ sin x cos y 4 cos x sin 2/ 

cos x cos 2/ 

_ sin (x + y) 



By [4], 

24. Prove that tan (x -f y) = 



cos x cos y 

sin 2 x + sin 2 2/ 



cos 2 x + cos 2 2/ 

By [20], sin 2 x + sin 2 2/ = 2 sin (x + y) cos (x — y). 

By [22], cos 2 x -f cos 2 y = 2 cos (x + 2/) cos (x — y). 

sin 2 x + sin 2 ?/ _ 2 sin (x 4- 2/) cos (x — ?/) 

cos 2 x + cos 2 2/ 2 cos (x + y) cos (x — 2/) 

= tan (x 4- 2/)- 

25. Prove that si ° X + C0S v = tan [ * ( * + y) + 45 ° ] ■ 

sin x — cos y tan [-J (x — y) — 45°] 

sin x 4- cos 2/ = sin x + sin (2/ 4- 90°) 
By [20], = 2 sin \ (x + y + 90°) cos i (x - 2/ - 90°). 

By [21], sin x - cos y - 2 cos \ (x -f 2/ 4 90°) sin \ (x - y - 90°). 

sin x 4- cos y _ tan £ (x 4- 2/ + 90°) 
* ' sin x — cos y tan ^ (x — y — 90°) 

= tan[i(s4-y) + 45 ] _ 
tan[£(x-2/)-45°]' 

26. Prove that sin 2 x + sin 4 x = 2 sin 3 x cos x. 
By [20], sin 2x 4- sin4x = 2 sin 3 x cos x. 

27. Prove that 

sin 4 x = 4 sin x cos x — 8 sin 3 x cos x = 8 cos 3 x sin x — 4 cos x sin x. 
By [12], sin 4x = 2 sin 2x cos 2x 

By [12] and [13], = 4 sin x cos x (1 -2sin 2 x) 

= 4 sin x cos x — 8 sin 3 x cos x. 
Again, sin 4 x = 4 sin x cos x (2 cos 2 x — 1) 

= 8 cos 3 x sin x — 4 sin x cos x. 

28. Prove that cos 4 x = 1 — 8 cos 2 x 4 8 cos 4 x = 1 — 8 sin 2 x 4 8 sin 4 x. 
By [13], cos4x = 2cos 2 2x- 1 

By [13], =2(2cos 2 x - l) 2 - 1 






teachers' edition. 



169 



Again, 
By [12], 



cos ix = 



8cos 4 x - 8cos 2 x 4- 2 - 

— 8 cos 2 x + 8 cos 4 x. 
-2sin 2 2x 

— 2 (4 sin 2 x cos 2 x) 

— 8 sin 2 a: (1 - sin 2 x) 

— 8sin 2 x -f 8sin 4 x. 



29. Prove that cos 2 x + cos 4 x = 2 cos 3 x cos x. 

By [22], cos 2 x + cos 4 x = 2 cos 3 x cos x. 

30. Prove that sin 3 x — sin x = 2 cos 2 x sin x. 
By [21], sin 3 x — sin x = 2 cos 2 x sin x. 

31. Prove that sin 3 x sin 3 x + cos 3 x cos 3 x = cos 3 2 x. 

sin 3 x sin 3 x = sin x sin 2 x sin 3 x 

= sin x (1 — cos 2 x) sin 3 x 
= sin x sin 3 x — sin x cos 2 x sin 3 x. 
cos 3 x cos 3 x = cos x cos 2 x cos 3 x 

= cos x (1 — sin 2 x) cos 3 x 
= cos x cos 3 x — cos x sin 2 x cos 3 x. 
.-. sin 3 x sin 3 x + cos 3 x cos 3 x 

= sin x sin 3 x + cos x cos 3 x 

— sin x cos 2 x sin 3 x — cos x sin 2 x cos 3 x 
By [9], =cos(3x-x) 

l — sin x cos x (cos x sin 3 x -f- sin x cos 3 x) 
By [4] , = cos 2 x — sin x cos x sin (3 x -f x) 

By [12], = cos2x — J sin 2xsin4x 

By [12], = cos2x — sin 2 2 x cos 2 x 

= cos 2 x (1 — sin 2 2 x) 
By [13], = cos 3 2x. 

32. Prove that cos 4 x — sin 4 x = cos 2 x. 

cos 4 x — sin 4 x = (cos 2 x + sin 2 x) (cos 2 x — sin 2 x) 
= 1 x cos2x 
== cos 2 x. 



33. Prove that cos 4 x -f sin 4 x = 1 — $ sin 2 2 x. 

cos 4 x + sin 4 x = (cos 2 x + sin 2 x) 2 — 2 sin 2 x cos 2 x 
= 1 — 2 sin 2 x cos 2 x 
= 1 --£sin 2 2x. 



170 



PLANE TRIGONOMETRY. 



34. Prove that cos 6 x — sin 6 x = (1 — sin 2 a: cos 2 x) cos 2x. 

cos 6 x — sin 6 x = (cos 2 x — sin 2 x) (cos 4 x -f cos 2 x sin 2 x + sin 4 x) 
By [13], = cos 2 x [(cos 2 x + sin 2 x) 2 — cos 2 x sin 2 x] 

= cos 2 x (1 — cos 2 x sin 2 x). 

35. Prove that cos 6 x -f sin 6 x = 1 — 3 sin 2 x cos 2 x. 

cos 6 x + sin 6 x ■= (cos 2 x -f sin 2 x) (cos 4 x — cos 2 x sin 2 a: + sin 4 x) 
= .cos 4 x — cos 2 x sin 2 x + sin 4 x 
= (cos 2 x + sin 2 a) 2 — 3 cos 2 x sin 2 x 
= 1 — 3 cos 2 x sin 2 x. 



r>^ -r. A , ■ sin 3 x + sin 5 x 

36. Prove that = cot x. 

cos3x — cos5x 

By [20] , sin 3 x + sin 5 X = 2 sin 4 x cos x. 

By [23] , cos 3 x — cos 5 x = 2 sin 4 x sin x. 

sin3x + sin5x cosx 



cos 3 x — cos 5 x sin x 



-= cotx. 



37. Prove that = 2 cos 2 x. 

sin x + sin 3 x 

By [20], sin3x + sin 5x = 2 sin 4 x cosx. 

By [20], sin x -f sin 3 x = 2 sin 2 x cos x. 

sin 3 x + sin 5 x sin 4 x 



By [12], 

38. Prove that esc x 
cscx 

By [12], 



By [13], 



sin x + sin 3 x 


sin 2x 




2 sin 2 x cos 2 x 




sin 2 x 




= 2 cos 2 x. 


esc x — 2 cot 2 x cos x = 2 sin x. 


— 2 cot 2 x cos x 


rt cos 2 x 

= esc x — 2 cos x 

sin2x 




2 cos 2 x cos x 

= CSC X — : 

2 sin x cos x 




1 cos 2 x 
sin x sin x 




1 — cos 2 x 




sinx 




2sin 2 x 


t • .- 


sinx 




= 2 sin x. 



teachers' edition. 171 

39. Prove that (sin 2 x — sin 2 y) tan (x -f y) = 2 (sin 2 x — sin 2 2/). 
By [21], sin 2x — sin 2 y = 2 cos(x + 2/)sin(x — 2/). 

.♦. (sin 2 x — sin 2 2/) tan (x + 2/)= 2 sin (x + ?/)sin(x — 2/). 
By [20], sin x + sin y = 2 sin \ (x + 2/) cos -J- (x — ?/). 

By [20], sin x — sin y = 2 cos -J (x + y) sin |(x — y). 

.-. sin 2 x — sin 2 ?/ = 4 sin £ (x + y) cos £ (x + y) sin £ (x ■*- y) cos £ (x - y) 
By [12], = sin (x + y) sin (x - y). 

.-. 2 (sin 2 x — sin 2 y) = 2 sin (x + 2/) sin (# — 2/) 

= (sin 2 x — sin 2 2/) tan (x + y) . 

40. Prove that (1 + cot x + tan x) (sin x — cos x) = 



csc 2 x sec 2 x 

w . v cos 2 x sin 2 x 

(l-f-cotx + tanx)(smx — cosx) = sinx — cosx + cosx : 1 sinx 

v /x ' sinx cosx 

sin 2 x cos 2 x 



cosx sinx 
secx cscx 



csc 2 x sec 2 x 

sin 2 3x 



41. Prove that sin x -f sin 3 x + sin 5 x = 

sinx 

By [20], sin x -f sin 5 x = 2 sin 3 x cos 2 x. 

.-. sin x + sin 3 x -f sin 5 x = sin 3 x -f 2 sin 3 x cos 2 x 

= sin 3 x (1 + 2 cos 2 x). 

By [21], sin 3x — sinx = 2 cos 2 x sinx. 

sin3x 

1 =.2 cos 2 x. 

sinx 
t sin 3 x 

1 -f .2 cos 2 x = • 

sin x i 

• o , • r v • o sin 3 as 

.-. sin x -f- sm 3 x + sin, 5 x ^ sin 3 x — — — 

sinx 

sin 2 3x 



>. « -^ ,, J cos x + cos 3 x 

42. Prove that = cot 3 x. 

3 sin x — sin 3 x 

3 cos x + cos 3 x = 2 cos x -f (cos x -f cos 3 x) 
By [22], = 2 cos x + 2 cos x cos 2 x 

= 2 cos x (1 + cos 2 x) 
By [17], = 4cos 3 x. 

3 sin x — sin 3 x = 2 sin x — (sin 3 ar— sin x) 



172 PLANE TRIGONOMETRY. 

By [21], = 2 sin x — 2 sin x cos 2 x 

= 2 sin x (1 — cos 2 x) 
By [16], = 4sin 3 x. 

3 cos x -f cos 3 x _ 4 cos 3 x 
' 3 sin x — sin 3 x 4 sin 3 x 
= cot 3 x. 

43. Prove that sin 3 x = 4 sin x sin (60° + x) sin (60° - x). 
By [4], sin (60° + x) = £ V3 cos x + i sin x. 
By [8], sin (60° - x) = £ V3 cos x - -J- sin x. 
.-. sin (60° + x) sin (60° - x) = f cos 2 x - I sin 2 x 

_ 3 (1 — sin 2 x) — sin 2 x 
~ 1 

_ 3 -4sin 2 x 
" 4 

4 sin x sin (60° + x) sin (60° - x) = sin x (3 - 4 sin 2 x) 

= 3sinx — 4sin 3 x 
By Prob. 18, Ex. XIV, = sin 3 x. 

44. Prove that sin 4 x = 2 sin x cos 3 x + sin 2 x. 
By [21], sin4x — sin2x = 2 cos 3 x sin x. 

.-. sin 4 x = 2 cos 3 x sin x -f sin 2 x. 

45. Prove that sin x -f sin (x — f it) 4- sin (J 7T — x) = 0. 
By [20], sin (x — § it) -f sin (± 7T — x) = 2 sin ( — \ it) cos (x — \ n) 

= — 2 sin J ^r sin x 
= — sin x. 
.*. sin x -f- sin (x — f at) -f sin (J ^ — x) = 0. 

46. Prove that cos x sin (y — z) + cos 2/ sin (z — x) -f cos z sin (x — y) = 0. 
By [8], cos x sin (y — z) = cos x sin y cos z — cos x cos ?/ sin z. 

cos y sin (z — x) = cos y sin z cos x — cos y cos z sin x. 
cos z sin (x — y) = cos z sin x cos y — cos z cos x sin y. 
.*. cos x sin (y — z) + cos y sin (z — x) -f cos z sin (x — y) = 0. 

47. Prove that 

cos (x -f y) sin i/ — cos (x + z) sin z = sin (x + y) cos y — sin (x + z) cos z. 

By [4] and [5], 

sin (x -f- y) cos y — cos (x + y) sin y — sin x (cos 2 ?/ + sin 2 2/) = sin x. 

sin (x -f z) cos z — cos (x -f- z) sin z = sin x (cos 2 z + sin 2 z) = sin x. 
.-. sin (x -f y) cos y — cos (x + y) sin y = sin (x + z) cos z — cos (x + z) sin z. 
.-. cos (x -f- y) sin 2/ — cos (x + z) sin z = sin (x -j- y) cos 2/ — sin (x + z) cos z. 



teachers' edition. 173 

48. Prove that 

cos (x 4- y 4- z) 4- cos (x + y - z) + cos (x - y 4- z) + cos (y + z - z) 

= 4 cos x cos y cos z. 

By [22], cos [(x + y) 4- 2] 4- cos [(x + ?/) - 2] - 2 cos (x + y) cos 2. 
cos [2 4- (x - ?/)] -f cos [2 — (x — y)] = 2 cos 2 cos (x — y). 
.-. cos (x 4- y 4- 2) -f cos (x + y - z) + cos (x - y + 2) 4- cos (2/ 4- 2 - x) 

= 2 cos (x + y) cos 2 4-2 cos (x — y) cos 2 

= 2 cos 2 [cos (x 4- y) 4- cos (x — y)] 

= 2 cos 2 (2 cos x cos y) 

— 4 cos x cos 2/ cos 2. 

49. Prove that sin (x 4- y) cos (x — y) 4- sin (y 4- 2) cos (?/ — 2) 

+ sin (2 4- x) cos (2 — x) = sin 2 x 4- sin 2 2/ 4- sin 2 2. 

By [20] , sin (x 4- y) cos (x — y) = i (sin 2 x 4- sin 2 y). 

sin (?/ 4- 2) cos (y — z) = i (sin 2 y + sin 2 2). 
sin (2 4- x) cos (2 — x) = \ (sin 2 2 4- sin 2 x). 
.*. sin (x 4- y) cos (x — y) 4- sin (y 4- 2) cos (y — 2) 4- sin (2 + x) cos (2 — x) 

= sin 2 x 4- sin 2 y 4- sin 2 2. 

^^ ^ xi- . sin 75° 4- sin 15° 

50. Prove that = tan 60°. 

sin 75° - sin 15° 

By [20], sin 75° 4- sin 15° = 2 sin 45° cos 30°. 

By [21], sin 75° - sin 15° = 2 cos 45° sin 30°. 

g sin 75° 4- sin 15° _ 2 sin 45° cos 30° 

' ' sin 75° - sin 15° ~" 2 cos 45° sin 30° 

= tan 45° cot 30° 

= tan 60°. 

51. Prove that cos 20° 4- cos 100° 4- cos 140° = 0. 

By [22], cos 20° 4- cos 100° = 2 cos 60° cos 40° 

= cos 40°. 
Also, cos 140° = cos (180° - 40°) 

= - cos 40°. 
.-. cos 20° 4- cos 100° 4- cos 140° = 0. 

52. Prove that cos 36° 4- sin 36° = V2 cos 9°. 

By Prob. 1, Ex. XXIV, cos 36° 4- sin 36° = V2 cos (36° - I it) 

= V2cos(-9°) 
= V2 cos 9°. 






174 PLANE TRIGONOMETRY. 

53. Prove that tan 11° 15' + 2 tan 22° 30' + 4 tan 45° = cot 11° 15'. 
By Prob. 11, Ex. XXIV, 

cot 11° 15' - tan 11° 15' = 2 cot 22° 30'. 
2 cot 22° 30' - 2 tan 22° 30' = 4 cot 45°. 
.-. cot 11° 15' - tan 11° 15' - 2 tan 22° 30' = 4 cot 45° = 4 tan 45°. 
.-. tan 11° 15' + 2 tan 22° 30' + 4 tan 45° = cot 11° 15'. 

54. If A, B, C are the angles of a plane triangle, prove that 

sin 2 A + sin 2 B + sin 2 C = 4 sin A sin B sin C. 

A + B+ C= 180°. 
By [20], sin 2 A + sin 2 £ = 2 sin (J. + 5) cos (A - £) 

= 2 sin O cos (A. - B). 
By [12], sin 2 C = 2 sin C cos C. 

.-. sin 2 A + sin 2 5 + sin 2 O = 2 sin C cos (J. - B) + 2 sin C cos (7 

= 2 sin C [cos (J. - B) - cos (A + B)] 
- By [23], =2sinC[-2sin./isin(-£)] 

= 4 sin C sin A sin 1?. 

55. If A, B, C are the angles of a plane triangle, prove that 

cos 2 A. + cos 2B + cos 2C = -l-4 cos A. cos 5 cos (7. 
By [22], cos 2 A + cos 2 £ = 2 cos (A. + B) cos (A - £) 

= -2cosCcos(A - B). 
.-. cos 2 A -j-cos2J5 + cos2C = -2cosCcos(A - B) + cos 2 C 
By [13], = -2cosCcos(A - J5) + 2 cos 2 C - 1 

= 2 cos C [cos C - cos(A - B)] - 1 

= 2 cos C [ - cos (A + B)- cos (A - B)] - 1 
By [22], = 2 cos C (- 2 cos A cos B) - 1 

= — 4 cos A cos 5 cos C — 1. 

56. If A, -B, C are the angles of a plane triangle, prove that 

■ n a • ot, • o^ „ 3 ^ 3 # 3Cf 

sin 3 J. + sm 3 B 4- sin 3C = -4 cos cos — cos 

2 2 2 

By [20], sin 3 A + sin 3 B - 2 sin § ( A + £)cosf (J. _ £) 

= 2 sin f (180° - C) cos | (A - B) 
= -2cosf Ccos§(A- J5). 
By [12], sin 3 C = 2 sin | O cos f (7 

= 2 sin f (180° - A - J5)cos§C 
= - 2 cos § (A + J5) cos f C. 
.-. sin3 A. + sinS-B + sin3 C = - 2 cos } C [cos f (A - B) + cos f (-4 + 5)] 

= - 2 cos f C (2 cos | A cos f £) 
= — 4 cos | A cos f J5 cos -f C. 



teachers' edition. 175 

57. If A, B, C are the angles of a plane triangle, prove that 
cos 2 A -f cos 2 B -f cos 2 (7 = 1—2 cos A cos B cos C. 

By [16], cos 2 4 = 

cos 2 £ = 

cos 2 C = 

2 

„ A . „ n . on 3 -f cos 2 4 + cos 2 £ + cos 2 C 

cos 2 A + cos 2 B -f cos 2 C = 

2 
By Prob. 55, Ex. XXIV, 

cos 2A-\- cos 2 B -f cos 2 C = — 1 — 4 cos 4. cos 5 cos G. 

3 — 1 — 4 cos A cos B cos (7 



1 + 


cos 


24. 


1 + 


2 
cos 


25 


1 + 


2 
cos 


2C 



,\ cos 2 A + cos 2 B + cos 2 (7 = 



2 
= 1 — 2 cos 4 cos B cos C. 



58. If A + £ + C = 90°, prove that 

tan A tan 5 + tan B tan C + tan C tan 4 = 1. 

tan 4 tan B -f tan 5 tan C + tan C tan 4 

= tan 4 tan B + (tan 4 -f tan 5) tan C 

tan J. + tan 5 

= tan A tan Z* H 

tan (A + t B) 

x» r^i a s. r. , tan4 + tan£ 

By [61, = tan J. tan B -\ 

L J tan ^4 + tan B 

1 — tan 4 tan B, 

= tan 4 tan B + 1 — tan 4 tan 5 

= 1. 



59. If A + JB + C = 90°, prove that 

sin 2 4 + sin 2 B + sin 2 C = 1 - 2 sin 4 sin E sin C. 

sin C = cos (A -f 5) 
By [5], = cos 4 cos B — sin A sin 5. 

sin C + sin A sin 5 = cos A cos 5. 
sin 2 C + 2 sin 4 sin J5 sin C + sin 2 A sin 2 5 
= cos 2 A cos 2 B. 
sin 2 C 4- 2 sin A sin 5 sin C = cos 2 4 cos 2 B - sin 2 J. sin 2 B 

= (1 - sin 2 A) (1 - sin 2 B) - sin 2 A sin 2 B 
= 1 -sin 2 A -sin 2 B. 
.-. sin 2 4 + sin 2 B -f . sin 2 C = 1 - 2 sin A sin 5 sin C 



176 



PLANE TRIGONOMETRY. 



60. If A + B + C = 90°, prove that 

sin 2 A 4- sin 2 B -f sin 2 (7 = 4 cos ^4 cos 2? cos (7. 

By [20], sin2^i + sin 2B = 2 sin(^L + B)cos(^L - B) 

= 2 cos cos (A - jB). 

By [12], sin 2 C = 2 sin C cos C. 

.-. sin 2 J. + sin 2 £ 4- sin 2 C = 2 cos C cos ( J. - B) + 2 sin (7 cos C 

= 2 cos C [cos ( J. - B) + sin C] 
= 2 cos C [cos (J. - B) + cos (^i 4- £)] 

By [22], = 4 cos -4 cos J5 cos O. 

61. Prove that sin (sin— l x 4- sin— l y) = x Vl — y 2 4- y Vl — x 2 . 
By [4], sin (sin- 1 x + sin 1 -^) = sin (sin-ix) cos (sin- 1 ^) 

4- cos (sin - 1 x) sin (sin~ l y) 
= x Vl - y2 + y Vl - x 2 . 

x4-y 



62. Prove that tan (tan— l x 4- tan- 2 y) = 
By [6] , tan (tan- l x 4- tan- 1 y) = 



1 -xy 
tan (tan- 1 x) 4- tan (tan— 1 y) 



63. Prove that 2 tan- l x = tan- 1 



x + y 
1 — tan (tan— l x) tan (tan— *y) 1 — xy 

2x 



By [14], 



tan (2 tan— 2 x) =± 
.*. 2 tan _1 x 



2 tan (tan— x x) 2x 



1 - 

tan -1 



tan 2 (tan- l x) 
2x 
1-x 2 ' 



64. Prove that 2 sin- 1 x = sin- 1 (2 x Vl - x 2 ) . 

By [12] , sin (2 sin- x x) = 2 sin (sin— x x) cos (sin— 1 x) = 2 x Vl — x 2 . 

.-. 2 sin- 1 x = sin- 1 (2 x Vl - x 2 ). 

65. Prove that 2 cos- x x = cos- 1 (2 x 2 - 1). 

By [13], cos (2 cos-ix) = 2 cos 2 (cos- l x) - 1 = 2x 2 - 1. 

.-. 2cos- x x = cos-i^x 2 - 1). 

3# _ x 3 

66. Prove that 3 tan- x x = tan- 1 — - • 

1 — 3x 2 

tan (3 tan _1 x) = tan (tan~ x x 4- 2 tan— 1 x) 

tan (tan— x x) 4- tan (2 tan- x x) 
1 — tan (tan— l x) tan (2 tan- 1 x) 
x 4- tan (2 tan— 1 x) 
1 — x tan (2 tan- 1 x) 



By [6], 



TEACHERS 7 EDITION. 

2x 



177 



z + 



By Prob. 63, Ex. XXIV, 



1-x 



.-. 3 tan-ix = tan— 1 



67. Prove that 
sin 



Let sin 
Then 



in- 1 */- = tan- 1 */ — 

i r* 

\/- = sum. 

\y 



f y - x 
\ y 

\y -x 



cosn. 



tan n. 
.-. n = tan— l 



\ y - x ' 
\y \y-x 



68. 


Prove that 


tan~ 




sin- 


\x - 


-y _ 

- z 


-l 


Let 






n. 




sin- 


'J- 


-y 





w 



Then a /- — - = sinn. 



\2/ - z 



cosn. 



tann. 
n = tan— 1 



\y 



-y 



..rin-iJ^^tan-iJEl. 

\x — z \?/ — 2; 

69. Prove that 1 

sin-ix = sec -1 - 



Vl -x 2 



1-x 2 _ 3 x - x 3 
2x ~l-3x 2 ' 
1 -x 2 
3x-x 3 
1 -3x 2 ' 



Let sin— 1 x = n. 

Then x = sin n. 

Vl — x 2 = cos n. 

1 

: = sec n. 



VT 



.-. sin-ix = sec - 1 



Vl _x 2 

1 



Vl -x 2 



70. Prove that 

2 sec - * x = tan - x 



2 Vx 2 - 1 



2 -x 2 
Let 2 sec— : x = n. 

Then x = sec £ n. 

1 



x 



cos £ n. 



By [13], 
ax 2 



2 ( - j — 1 = cos n. 



2-x 2 



= cos n. 



2-x 2 



= sec n. 



By Prob. 2, Ex. V, 
/_&^_\ 2 _ 1 2 

\2-x 2 / 



4x 2 -4 

(2 - x 2 ) 2 



= tan 2 n. 



tan n ■■ 



2 Vx 2 - 1 



2-x 2 



n = tan~ 



, 2 Vx 2 - 1 
2-x 2 



.-. 2 sec- 1 x = tan- 1 



2 Vx 2 - 1 
2-x 2 



178 PLANE TRIGONOMETRY. 

71. Prove that tan- 1 1 + tan- 1 f = 45°. 






By [6], tan (tan- 1 § + tan- 1 i) = 2 "^ 3 i = 1. 

1 ~~ 2 X 3 

.-. tan- 1 i + tan- 1 1 = tan- 1 1 = 45°. 



72. Prove that tan- 1 1 -f tan- 1 \ = tan- x f 
By [6] , tan (tan- * f + tan- 1 1) = 



+ ^ _, 



.-. tan- 1 i -f tan— 2 1 = tan— 1 $ . 

73. Prove that sin- 1 1 -f- sin- 1 if = sin- 1 ff . 
By [4] , sin (sin- * f + sin- 1 if) 

== sin (sin— 1 §) cos (sin~ l if) -f cos (sin- 1 f ) sin (sin— 1 if) 

— 3 v _5 J. 4 y 12 - 6 3 

.-. sin— x f + sin -1 1| = sin -1 £§. 

1 4 

74. Prove that sin— 1 — — + sin -1 = 45°. 

V82 Vil 

V82 Vil/ V82 Vil V82"Vil 

5 + ^=^=*^ 



By [4], sin ( sin- 1 — — + sin- 1 —— ) = —— x -— + — = x 
V VS2 V41 / V82 Vil 



41 V2 41 V2 V2 

14/- 
.-. sin- 1 h sin- 1 = sin-H V2 = 45°. 

V82 Vil 

75. Prove that sec- 1 f + sec- 1 if = 75° 45'. 

sec- 1 f + sec- 1 |f = cos- x § -f- cos - 2 |f. 
By [6], cosmos- 1 ! + cos-itf) = f X if - J x A = it- 

.*. cos -1 J + cos -1 if == cos -1 Jf. 

sec- 1 f + sec- 1 if = sec- 1 } £ = 75° 45'. 

76. Prove that tan- 1 (2 + Vs) - tan- 1 (2 - V3) = sec™ 1 2. 
Let tan- 1 (2 + V3) - tan- 1 (2 - Vs) = n. 

(2 + V§) - (2 - VS) 
By [10], tan n = A_Z L^ ^_ 

1 + (2 + V3) (2 - V3) 

2 V3 



2 

.-. n = 60°. 
sec n = 2. 
.-. tan- 1 (2 + Vs) - tan-- 1 (2 - VS) = sec- 1 2. 



Vs. 






teachers' edition. 179 

77. Prove that tan- 1 i + tan- 1 i + tan- 1 f + tan- 1 \ = 45°. 
Let tan- 1 J + tan- 1 i = n, 

and tan- 1 J + tan- 1 J = ». 

By [6], tann = T -i±i- T = f 



tan u = 



4 4-1 

T JL 8 _ _3_ 



l-fxi 
tan (n + v) = * + * = 1. 

1 T X TT 

?i + v = tan-il = 45°. 
tan- 1 J + tan- 1 £ + tan- 1 } + tan- 1 j = 45°. 

78. Prove that tan- : + tan- 1 — - = tan- 1 — — - 

l-2x + 4x 2 l+2x + 4x 2 2x 2 

By [6] , tan ( tan- x — + tan~ * — ) 

J L J ' V l-2x + 4x 2 l + 2x + 4x 2 / 







1 , 


1 




1-2 


x + 4x 2 ' 1 + 2 


x + 4x 2 


1 




1 






(1- 

1+2 


-2x + 4x 2 )(l + 
x + 4x 2 + 1 -2 


2x + 4x 2 ) 
x + 4x 2 



(1 -2x + 4x 2 )(l + 2x + 4x 2 )-l 
2 + 8x 2 
~4x 2 + 16x 4 

1 
"2X 2 ' 

.-. tan- 1 h tan -1 = tan- 1 

l_2x + 4x 2 l+2x + 4x 2 2x 2 

79. Given cos x = f ; find sin J x and cos J x. 



By [16], sin| !C = ± ^Q = ±V7 = ±t V6- 

By [17], C osix = ±^l±i = ±V| = ±sV5. 

80. Given tan x = \ ; find tan J x. 
By [14], tanx: 



1 - tanHx 



1 - tan 2 £x 



180 PLANE TRIGONOMETRY. 



1 — tan 2 J x = 4 tan \ x. 



tan 2 \ x 4- 4 tan § x = 1. 
tan 2 Jx + 4 tan |x + 4 = 5. 

tan-|x + 2 =±V6. 
.*. tan \x = ± V5 - 



2* 



81. Given sin x + cos x = Vi ; find cos 2 x. 
sin x + cos x = VI 
sin 2 x + 2 sin x cos x 4- cos 2 x = \. 
1 4- 2 sin x cos x = J. 
2 sin x cos x = — J. 
By [12], sin2x = -i 



cos2x = ±Vl -(-i) 2 

:±Vf = ±jV3. 



82. Given tan 2 x = - 2 / ; find sin x. 

tan 2 x ss - 2 7 4 -. 
By Prob. 19, Ex. XIV, sec 2 2 x = 1 + tan 2 2 x 



cos 2 2x = ^V 
cos 2 x = ± 2V 
By [13], l-2sin 2 x = ±2V 

sin 2 x = 2% or if. 
.-. sin x = ± I or ± f . 

83. Given cos 3 x = § f- ; find tan x. 
By Prob. 19, Ex. XIV, cos 3 x = 4 cos 3 x — 3 cos x. 
4 cos 3 x — 3 cos x = ff. 
108 cos 3 x - 81 cos x = 23. 
108 cos 3 x - 81 cos x - 23 = 0. 
(3 cos x + 1) (36 cos 2 x - 12 cos x - 23) = 0. 
(i) 3 cos x + 1 = 0. 

cos x = — J. 
(ii) 36 cos 2 x - 12 cos x - 23 = 0. 

36 cos 2 x - 12 cos x 4- 1 = 24. 

6 cos x - 1 = ± 2 V6. 

cosx = i(l ± 2 V6). 
lX = ± VT^ 



sinx = ±vi — cos^x 



:4-Vl-(-i)2 o r ± Vl -[£ (1 ±2V6) 2 
= ±V| or ±V^(11 T 4V6) 
= ±fV5 or ± i(V3qp2V2). 



TEACHERS' EDITION. 



181 



tanx = 



= =F2 V2 



|V5 i:i(V3=F2 V2) 
1 i(l±2V6) 

± (V3 q=2 V2) 



1 ±2 V6 



/- ±(3-2V2)(l-2V6) j:(V3 + 2V2)(l + 2V6) 

— t ^ v *i °r 7= t=~ j or — — — 

(1 + 2 V6) (1 - 2 V6) (1 - 2 V6) (1+ 2 V6) 



23 



23 



.-. tanx = ± 2 V2, ± ^(9 V3 + 8 V2), or ± JL (9 V3 - 8 V2). 



84. Given 


2 esc x - 


- cot x = v3 


find sin ^ x. 






2 esc x - 


- cot x = 


V3. 


2 


cosx 


V3. 


smx 


sin x 




2 - 


- cos x = 


V3 sin x. 


4 — 4 cos x -f 


cos 2 x = 


3sin 2 x 




= 


3-3cos 2 x. 


4cos 2 x — 4cosx + 1 = 


0. 


2 cos x — 1 = 


0. 




cosx = 
sin i x = 


1 

2* 


By [16], 


,.p. 


1 \ 2 


.-. sin \x = 


±s- 



85. Find sin 18° and cos 36°. 
(i) 54° = 90° - 36°. 

3 x 18° = 90° - 2 x 18°. 
cos (3 x 18°) = sin (2 x 18°). 
By Prob. 19, Ex. XIV, and [12], 
4 cos 3 18° -3 cos 18° 

= 2 sin 18° cos 18°. 
4 cos 2 18° - 3 = 2 sin 18°. 
4-4 sin 2 18° - 3 = 2 sin 18°. 
4 sin 2 18° + 2 sin 18° = 1. 
16sin 2 18° + () + l = 5. 

4 sin 18° + 1 = ± Vs. 
4 sin 18° = ± V5 - 1. 



.-. sin 18° 



V5- 1 



(ii) 



cos 36° = 1 - 2 sin 2 18° 

V5 + 1 



86. Find the value of 
a sec x -f b esc x, when tan x = 

tan x = — • 
a 5 



3/6 
a" 



By Prob. 2, Ex. V, 
sec 2 x = 1.+ 



6* 



+ 6* 









ai 






cotx = 




By 


Prob 


3, Ex. 


V, 






CSC 2 X = 


a* + 6* 




•* 


. sec x = 

CSC x = 


(a* + bi)i 
_ (a* + 6*)* 



6* 



182 



PLANE TRIGONOMETRY. 



a sec x -f b esc x 

= cfi (a* + b*)* + b$ (a* + &*)* 
= (a* + &*) (a* + M)* 
= (a* + & f ) f . 



87. Find the value of sin 3 a, 
when sin 2 x = Vl — ra 2 . 



sin 2 x = Vl — m 2 , 
cos 2 2 x = m 2 . 
cos 2 x = ± m. 
1—2 sin 2 x = ± m. 
2sin 2 x = 1 ± ra. 



->E 



sinx 



By Prob. 18, Ex. XIV, 

sin 3 x = 3 sin x — 4 sin 3 x 
'1 ± ra\£ 



± ra 



)--*(4=) 



3-4 



2 



= 3( 

(1 ± wi\i^ rt v 
— )(lT2m). 



88. Find the value of sin x, when 
tan 2 x + 3cot 2 x = 4. 

tan 2 x + 3 cot 2 x = 4. 
3 



tan 2 x + ■ 



= 4. 



tan 2 x 

tan 4 x - 4 tan 2 x + 3 = 0. 
(tan 2 x - 1) (tan 2 x - 3) = 0. 

.-. tan 2 x = 1 or 3. 

cot 2 x = 1 or J. 

csc 2 x = 2 or f. 

sin 2 x = \ or f . 

/. sinx = ± |V2 

or ± ^V3. 



89. Find the value of 
csc 2 x — sec 2 x 






csc 2 x + sec 2 x 



, when tan x : 



:VT. 



= VT. 



tanx 

sec 2 x = 1 + \ 

cot x = V7. 

csc 2 x = 1 + 7 
csc 2 x — sec 2 x _ 8 — f 
csc 2 x-fsec 2 x 8 + f 



8. 



= *. 



90. Find the value of cos x, when 
5 tan x -f sec x = 5. 

5 tan x + sec x = 5. 



5 sinx 



• + • 



5. 



25(1 



COS X COS X 

5 sin x -f 1 = 5 cos x. 
5 sinx = 5cosx — 1. 
cos 2 x) 

= 25 cos 2 x — 10 cos x -f 1. 
50 cos 2 x - 10 cos x- 24 = 0. 
25cos 2 x — 5 cos x — 12 = 0. 
(5cosx -4)(5cosx + 3) = 0. 

5 cos x = 4 or — 3. 
.-. cosx = f or — f. 



91. Find the value of sec x, when 

a 
tanx : 



V2a 


+ 1 


tan x 


a 




V2a + 1 


sec 2 x 


-11 ^ 


' 2a + l 




a 2 + 2 a + 1 




2a + 1 


secx 


a + 1 



V2a+ 1 



92. Simplify 



TEACHEKS' EDITION. 183 

(cos x + cos y)' 2 + (sin x 4- sin y)' 2 



cos 2 1 (x — y) 
By [22] and [20], 

(cos x + cos y) 2 4- (sin x + sin y) 2 
cos 2 ^ (x — y) 
_ [2 cos j (x 4- y) cos | (x - y)] 2 + [2 sin j (x 4- y) cos j (x - y)] 2 

COS 2 J- (X — y) 

= 4 cos 2 i(x + y) 4- 4 sin 2 £(x 4- y) 
= 4. 

93. Simplify Sin(X + 2y) - 2sin(X + y) + SinX . 

cos (x + 2 y) — 2 cos (x 4- y) 4- cos x 

sin (x 4- 2 y) — 2 sin (x 4- y) 4- sin x 
cos (x 4- 2 ?/) — 2 cos (x + y) -f cos x 

_ [sin (x 4- 2 ?/) 4- sin x] — 2 sin (x 4- 2/) 
[cos(x 4- 2 ?/) 4- cosx] — 2cos(x 4- y) 
By [20] and [22], = 2Bin (x + y )co» y - 2 8in(x + y) 
2 cos (x 4- y) cos y — 2 cos (x 4- y) 
_ sin (x 4- y) (cos ?/ — 1) 
cos (x 4- y) (cos y — 1) 
_ sin (x 4- y) 
cos (x 4- ?/) 
= tan (x + y). 

ft* o- !•* sin (x - 2:) 4- 2 sin x 4- sin (x 4- 2;) 

94. Simplify 

sin (y - z) 4- 2 sin y 4- sin (y 4- 2) 

sin (x — z) 4- 2 sin x 4- sin (x 4- 2) 
sin (y — z) 4- 2 sin y 4- sin (?/ 4- 2) 

_ [sin (x 4- z) 4- sin (x — z)] 4- 2 sin x 



By [20], 



[sin (y 4- z) 4- sin (?/ - 2)] + 2 sin */ 
_ 2 sin x cos z 4- 2 sin x 

2 sin y cos z 4- 2 sin y 
_ sin x (cos 2 4- 1) 

sin 2/ (cos 2 + 1) 
_ sinx 

sin y 

0.1- r,. ,., cos 6 x — cos 4 x 

95. Simplify 

sin 6 x 4- sin 4 x 

By [23] and [20], 

cos 6 x — cos 4 x _ — 2 sin 5 x sin x _ — sin x 

sin 6 x + sin 4 x 2 sin 5 x cos x cos x 



= — tan x» 



184 PLANE TRIGONOMETRY. 

96. Simplify tan- 1 (2 x + 1) + tan- i (2 x - 1). 
tan [tan- 1 (2 x + 1) + tan-!(2 x - 1)] 

2x + 1 + 2x - 1 



By [6], 



l-(2x + l)(2x- 1) 

4x 
2-4x 2 

2x 



1 -2x 2 
;•■ tan-!(2£ + 1) + tan- 1(2 » - 1) = tan- 1 



2x 



1 -2x 2 



97. Simplify 3L_ + 1 + 



1 + sin 2 x 1 -f cos 2 x 1 + sec 2 x 1 -f csc 2 x 
1 1 1 1 

1 + sin 2 x 1 -f cos 2 x l + sec 2 x 1 + csc 2 x 

Vl + sin 2 x 1 -f csc 2 x/ \1 + cos 2 x 1 -f sec 2 x/ 

(1 sin 2 x \ / 1 cos 2 x \ 

1 -f- sin 2 x 1 + sin 2 x/ \1 + cos 2 x 1 + cos 2 x/ 

_ 1 + sin 2 x 1 + cos 2 x 

1 + sin 2 x 1 + cos 2 x 
= 1 + 1 
= 2. 

98. Simplify 2 sec 2 x — sec 4 x — 2 csc 2 x 4- csc 4 x. 

2 sec 2 x — sec 4 x — 2 csc 2 x -f csc 4 x 

= 1 — 2 csc 2 x + csc 4 x — 1 + 2 sec 2 x — sec 4 x 
= (csc 2 x - l) 2 - (sec 2 x - l) 2 
== cot 2 x — tan 2 x. 



99. Solve sin x = 2 sin (J it + x). 

sin x = 2 sin (-J- it + x) 
By [4], =2 sin \ it cos x -j- 2 cos J ^ sin x 

= V3 cos x -f sin x. 
V3 cos x = 0. 
cos x = 0. 
.-. x = 1 7T or f 7T. 

100. Solve sin 2 x = 2 cos x. 

sin 2 x = 2 cos x. 
2 sin x cos x = 2 cos x. 
2 cos x (sin x — 1) = 0. 









teachers' edition. 185 



(i) 2 cos x = 0. 

cos x = 0. 
.-. x = 90° or 270°. 
(ii) sin x — 1 = 0. 

sin x = 1. 
.-. x = 90°. 
.-. x = 90° or 270°. 

101. Solve cos 2 x = 2 sin x. 







cos2x 


= 2 


sinx 






1 


— 2sin 2 x 


= 2 


sinx 






2 sin 2 x 


+ 2 sin x 


= 1. 






4 


sin 2 x + 4 


sin x + 1 


= 3. 








2 


sin x -h 1 


= ± 


Vs. 








2 sinx 


= — 


i± 


yfe. 






sinx 


= ~ 


i± 

2 


V3 










1± 


1.7320 



2 
= 0.3660 or -1.3660. 
.-. x = 21° 28 r or 158° 32'. 

102. Solve sin x + cos x = 1. 

sin x + cos x = 1. 

sin 2 x + 2 sin x cos x -f cos 2 x = 1. 

2 sin x cos x = 0. 

(i) sin x = 0. 

.-. x = 0° or 180°. 

(ii) cos x = 0. 

.-. x = 90° or 270°. 

.-. x = 0°, 90°, 180°, or 270°. 
But the values x = 180° and x = 270° do not satisfy the given equation. 

.-. x = 0° or 90°. 

| 

103. Solve sin x + cos 2 x = 4 sin 2 x. 

sin x -f cos 2 x = 4 sin 2 x. 
sin x + 1 — 2 sin 2 x = 4 sin 2 x. 
6 sin 2 x — sin x — 1 = 0. 
(2 sin x - 1) (3 sin x + 1) = 0. 



186 PLANE TRIGONOMETRY. 

(i) 2 sin x - 1 = 0. 

2 sin x == 1. 
sin x = J. 
.-. x - 30° or 150°. 

(ii) 3 sin x + 1 = 0. 

3sinx = — 1. 
sin x = — J. 
.-. x = 199° 28' or 340° 32'. 
.-. x = 30°, 150°, 199° 28', or 340° 32'. 

104. Solve 4 cos 2 x + 3 cos x = 1. 

4 cos 2x + 3 cos x = 1. 

8 cos 2 x — 44-3 cos x = 1. 

8 cos 2 x 4- 3 cos x — 5 = 0. 

(cos x 4- 1) (8 cos x — 5) = 0. 

(i) cos x 4- 1 = 0. 

cos x = — 1. 
.-. x = 180°. 

(ii) 8 cos x — 5 = 0. 

8 cos x = 5. 
cosx = | = 0.6250. 
.-. x = 51° 19' or 308° 4K 
.-. x = 51° 19', 180°, or 308° 41". 

105. Solve sin x 4- sin 2 x = sin 3 x. 

sin x 4- sin 2 x = sin 3 x. 
By [12] and Prob. 18, Ex. XIV, 

sin x 4- 2 sin x cos x = 3 sin x — 4 sin 8 x. 

4 sin 3 x — 2 sin x 4- 2 sin x cos x = 0. 

sin x (2 sin 2 x — 1 4- cos x) = 0. 

sin x (1 — 2 cos 2 x 4- cos x) = 0. 

sin x (1 — cos x) (1 4- 2 cos x) = 0. 

(i) sin x = 0. 

.-. x = 0° or 180°. 

(ii) 1 — cos x = 0. 

cosx = 1. 
.♦. x = 0°. 

(iii) 1 4- 2 cos x = 0. 

2 cosx = — 1. 
cos x = — J-. 
,\x = 120° or 240°. 
.-. x = 0°, 120°, 180°, or 240°. 



187 



106. Solve sin 2 x = 3 sin 2 x — cos 2 x. 

sin 2 x = 3 sin 2 x — cos 2 x. 
By [12], 2 sin x cos x = 3 sin 2 x — cos 2 x. 

3 sin 2 x — 2 sin x cos x — cos 2 x = 0. 
(3 sin x -f cos x) (sin x — cos x) = 0. 
(i) 3 sin x + cos x = 0. 

3 tan x + 1 = 0. 
tan x = — i. 
.-. x = tan-i (- i) = 161° 34' or 341° 34'. 
(ii) sin x — cos x = 0. 

sin x = cos x. 



.-. x = 45° or 225°. 

.-. x = 45°, 161° 34', 225°, or 341° 34'. 



107. Solve cot = i tan 0. 



cot = | tan 0. 
1 tan0 



tan 3 
tan 2 (9 = 3. 
tan = ± Vs. 
.-. = 60°, 120°, 240°, or 300°. 

108. Solve 2 sin = cos 0. 

2 sin = cos 0. 

4 sin 2 = cos 2 0. 

4 sin 2 = 1- sin 2 0. 

5sin 2 = 1. 

sin 2 = i 

sin = ± I V5 = ± 0.4472. 

.-. = 26° 34', 153° 26 y , 206° 34', or 333° 26'. 

But the values = 153° 26' and = 333° 26' do not satisfy the given 

equation. 

.-. = 26° 34' or 206° 34'. 

109. Solve 2 sin 2 x 4- 5 sin x = 3. 

2 sin 2 x + 5 sin x = 3. 

2 sin 2 x + 5 sin x — 3 = 0. 

(sin x + 3) (2 sin x - 1) = 0. 

(i) sin x + 3 = 0. 

sin x = — 3. 
.-. x is impossible. 






188 PLANE TRIGONOMETRY. 

(ii) 2 sin x - 1 = 0. 

2sinx = 1. 
sin x = i. 
.-. x = 30° or 150°. 

110. Solve tan x sec x = V2. 

tan x sec x = V2. 
tan 2 x sec 2 x = 2. 
By Prob. 2, Ex. V, 

tan 2 x(l + tan 2 x) = 2. 

tan 2 x + tan 4 x = 2. 

tan 4 x + tan 2 x — 2 = 0. 

(tan 2 x - 1) (tan 2 x + 2) = 0. 

(i) tan 2 x-l=0. 

tan 2 x = 1. 
tan x = ± 1. 

.-. x = 45°, 135°, 225°, or 315°. 

(ii) tan 2 x + 2 = 0. 

tan 2 x = - 2. 
tanx = d=V-2. 
.•. x is impossible. 
.-. x = 45°, 135°, 225°, or 315°. 
But the values x = 225° and x = 315° do not satisfy the given equation. 

.-. x = 45° or 135°. 

111. Solve sin x = cos 2 x. 

sin x = cos 2 x. 
By [13], sin x = cos 2 x — sin 2 x. 

sin x = 1 — sin 2 x — sin 2 x. 
2 sin 2 x + sin x — 1 = 0. 
(sin x + 1) (2 sin x - 1) = 0. 

(i) sin x + 1 = 0. 

sin x = — 1. 
.-. x = 270°. 

(ii) 2 sin x - 1 = 0. 

2 sin x s= 1. 
sin x = ■}-. 
.-. x ■= 30° or 150°. 
.-. x = 30°, 150°, or 270°. 



teachers' edition. 189 



112. Solve tan x tan 2 x = 2. 

tan x tan 2 x — 2. 

2 tan x 

By ri41, tanx ^- = 2. 

J L J 1 -tan 2 x 



2 tan 2 x - 2 - 2 tan 2 x. 
4tan 2 x = 2. 
tan 2 x = £. 

tanx=±Vi = ±!V2=± 0.7071. 
•. x = 35° 16', 144° 44', 215° 16', or 324° 44'. 



113. Solve sec x = 4 esc x. 



sec x = 4 esc x. 
1 4 



cos x sin x 
sin x = 4 cos x. 
sin 2 x = 16 cos 2 x. 
1 — cos 2 x = 16 cos 2 x. 
17 cos 2 x = 1. 

cos x=± T \ Vl7 = ± 0.2425. 
.-. x = 75° 58', 104° 2', 255° 58', or 284° 2'. 
But the values x = 104° 2' and x = 284° 2' do not satisfy the given 
equation. 

.-. x = 75° 58' or 255° 58'. 



114. Solve cos + cos 2 = 0. 

cos + cos 2 = 0. 

By [13], cos + cos 2 - sin 2 0=0. 

cos + cos 2 - 1 -f cos 2 = 0. 

2 cos 2 + cos - 1 = 0. 

(cos + 1) (2 cos - 1) = 0. 

(i) cos + 1 = 0. 

cos = — 1. 
.-. = 180°. 

(ii) 2 cos - 1 = 0. 

2 cos = 1. 
cos = -J. 
.-. = 60° or 300°. 
.-. = 60°, 180°, or 300°. 



190 PLANE TRIGONOMETRY. 

115. Solve cot i + esc = 2. 

cot£0 -f csc0 = 2. 

By [19], ± Ji_±^ + JL 2 . 

\ 1 — cos sin 



^ 



COS0 

A — • 



cos sin 

1 + cos 4 1 

= 4 J 

1 — cos sin sin 2 

1 -f 2 cos (9 + cos 2 A 4 1 

— : 4 — 



l-cos 2 sin0 sin 2 

1 + 2 cos + cos 2 (9,4 1 

-= 4 — 



sin 2 sin0 sin 2 

t. * * 1 + cos 1 

Extract square root, = 2 

sin sin 

1 -{- cos0 = 2sin0 — 1. 

2 + cos = 2 sin_0. 

2 + cos = 2 Vl - cos 2 0. 

4 + 4 cos + cos 2 — 4-4 cos 2 0. 
5 cos 2 + 4 cos = 0. 
cos 0(5 cos + 4) = 0. 

.\ cos0 = O or -0.8. 
Also 2 + Vl - sin 2 = 2 sin 0. 

1 - sin 2 = 4 sin 2 - 8 sin + 4. 
5 sin 2 - 8 sin + 3 = 0. 
(sin - 1) (5 sin - 3) = 0. 

.-. sin = 1 or 0.6. 
But cos0 = O or -0.8. 

.-. = 90° or 143° 8'. 

116. Solve cot x tan 2 x = 3. 

cot x tan 2 x = 3. 

_ _.,._ 1 2tanx _ 

By [14], x = 3. 

tan x 1 — tan 2 x 

2 =3. 



— tan 2 x 

2 = 3 -3tan 2 x. 
3tan 2 x = 1. 
tan 2 x = -J-. 

tan x • = ± V^ = ± £ Vs. 
.-. x = 30°, 150°, 210°, or 330°. 



teachers' edition. 191 



117. Solve sin x sec 2x^1. 

sin x sec 2 x = 1. 

sin x _ 



cos2x 

sin x = cos 2 x. 
By [13], sin x = cos 2 x — sin 2 x. 

sin x = 1 — 2 sin 2 x. 
2 sin 2 x + sin x — 1 = 0. 
(sin x -f 1) (2 sin x - 1) = 0. 

(i) sin x + 1 = 0. 

sin x = — 1. 
.-. x = 270°. 

(ii) 2 sin x - 1 = 0. 

2sinx = 1. 
sin x — \. 
.-. x = 30° or 150°. 
.\ x = 30°, 150°, or 270°. 



118. Solve sin 2 x + sin 2 x = 1. 

sin 2 x + sin 2x = 1. 

By [12], sin 2 x -f- 2 sin x cos x = 1. 

sin 2 x + 2 sin x vi — sin 2 x = 1. 

2 sin x Vl — sin 2 x = 1 — sin 2 x. 
4 sin 2 x — 4 sin 4 x = 1 — 2 sin 2 x + sin 4 x. 
5 sin 4 x — 6 sin 2 x + 1 = 0. 
(sin 2 x - 1) (5 sin 2 x - 1) = 0. 

(i) sin 2 x-l = 0. 

sin 2 x = 1. 
sinx = ± 1. 
.-. x = 90° or 270°. 

(ii) 5sin 2 x-l=0. 

5sin 2 x = 1. 
sin 2 x — \. 

sinx = ±V^ = ±iV5 = ± 0.4472. 
.-. x = 26° 34 r , 153° 26', 206° 34', or 333° 26'. 
;•. x = 26° 34', 90°, 153° 26', 206° 34', 270°, or 333° 26'. 
But the values x = 153° 26' and x = 333° 26' do not satisfy the given 
equation. 

.-. x = 26° 34', 90°, 206° 34', or 270°. 



192 PLANE TRIGONOMETRY. 

119. Solve cos x sin 2 x esc x = 1. 

cos x sin 2 x esc x = 1. 

By [12], cos x(2 sin x cos x) x = 1. 

sinx 

2 cos 2 x = 1. 
cos 2 x = -J. 

cos x = ± Vj = ± -J V2. 
.-. x = 45°, 135°, 225°, or 315°. 

120. Solve cot x tan 2 x = sec 2 x. 



By [14] and [13], 



cot x tan 2 x = sec 2 x. 
1 2tanx 1 



tanx 1 — tan 2 x cos 2 x — sin 2 x 
2 1 



1 — tan 2 x cos 2 x — sin 2 x 
sin 2 x 



2 cos 2 x — 2 sin 2 x = 1 

2 -2sin 2 x- 2sin 2 x= 1 

1—4 sin 2 x = - 



cos 2 x 

sin 2 x 
1 — sin 2 x 
^in 2 x 



1 — sin 2 x 

1 — 5 sin 2 x + 4 sin 4 x = — sin 2 x. 

1-4 sin 2 x -f 4 sin 4 x = 0. 

Extract root, 1—2 sin 2 x — 0. 

2sin 2 x= 1. 

sin 2 x = •}. 

sin x = ± V| = ± £ V5. 
.♦. x = 45°, 135°, 225°, or 315°. 

121. Solve sin 2 x = cos 4 x. 

sin 2 x = cos 4 x. 
By [13] , sin 2 x = cos 2 2 x - sin 2 2 x. 

sin 2 x = 1 — sin 2 2 x — sin 2 2 x. 
2 sin 2 2 x + sin 2 x - 1 = 0. 
(sin2 x + 1) (2 sin 2 x - 1) = 0. 

(i) sin 2 x + 1 = 0. 

sin 2 x = — 1. 
.-. 2 x = 270° or 630°. 
./. x = 135° or 315°. 



teachers' edition 193 

(ii) 2 sin 2 x - 1 = 0. 

2 sin 2 x = 1. 
sin 2 x = i. 
.-. 2 x = 30°, 150°, 390°, or 510°. 
.-. x = 15°, 75°, 195°, or 255°. 
.-. x = 15°, 75°, 135°, 195°, or 255°. 

122. Solve sin 2 z cot z — sin 2 z = £. 

sin 2 z cot 2 — sin 2 z = \. 

By [12], 

« • COS 2 . 

2 sin 2 cos z x sm 2 z = §. 

sin 2 

2 cos 2 z — sin 2 g = i. 

2 — 2 sin 2 2 — sin 2 z — \. 

3 sin 2 z — § . 

sin 2 g = £. 

sin 2 = ± V| = ± i V£ 

.-. 2 = 45°, 135°, 225°, or 315°. 

123. Solve tan x + tan 2 x = tan 3 z. 

tan x + tan 2 x = tan 3 x. 

By [14], and Prob. 19, Ex. XXIV, 

2 tan x 3 tan x — tan 3 x 
tanx + 



tan x( 1 4- 



1 — tan 2 x 1 — 3 tan 2 x 

tan 2 x 



1 — tan 2 x 1 — 3 tan 2 x 



2 x/ 



(i) tan x = 0. 

.-. x = 0° or 180°. 

gi) 14— J 3 - tan ' a =0. 

1 -tan 2 x 1 -3tan 2 x 

1 - 4 tan 2 x 4- 3 tan 4 x 4-2-6 tan 2 x - 3 4- 4tan 2 x - tan 4 x + = 0. 

2 tan 4 x -6tan 2 x = 0. 

tan 2 x (tan 2 x - 3) = 0. 

tan 2 a; = or 3. 

tan x = 0. 

.-. x = 0° or_180°. 

tan x = ± V3. 

.-. x = 60°, 120°, 240°, or 300°. 

.-. x = 0°, 60°, 120°, 180°, 240°, or 300°. 



194 PLANE TRIGONOMETRY. 

124. Solve cot x 



anx = sin x+ cos x. 




cot x — tan x = 


= sin x -j- cos x. 


cos x sin x 


- sin x -f cos x. 


sin x cos x 


cos 2 x — sin 2 x 


= sin x 4- cos x. 



sin x cos x 

cos 2 x — sin 2 x = sin x cos x (sin x -f cos x). 
(sin x 4- cos x) (cos x — sin x — sin x cos x) = 0. 

(i) sin x + cos x = 0. 

sin x = — cos x. 
.-. x = 135° or 315°. 

(ii) cos x — sin x — sin x cos x = 0. 

cos x — sin x = sin x cos x. 
Square, cos 2 x -f sin 2 x — 2 sin x cos x = sin 2 x cos 2 x. 

1 — 2 sin x cos x = sin 2 x cos 2 x. 
sin 2 x cos 2 x + 2 sin x cos x = 1. 
sin 2 x cos 2 x + 2 sin x cos x + 1 = 2. 
Extract the root, sin x cos x -f 1 = ± V2. 

sin x cos.x = — 1 ± V2. 
2 sin x cos x = — 2 ± 2 V2. 
By [12], sin 2 x = - 2 ± 2 V2. 

sin 2 x = 0.8284 or - 4.8284. 
.\ 2 x = 55° 56", 124° 4', 415° 56', or 484° 4'. 
.-. x = 27° 58', 62° 2', 207° 58 r , or 242° 2'. 
.-. x = 27° 58 7 , 62° 2', 135°, 207° 58', 242° 2', or 315°. 
But the values x = 62° 2' and x = 207° 58 x do not satisfy the given 
equation. 

.-. x = 27° 58 7 , 135°, 242° 2', or 315°. 

125. Solve tan 2 x = sin 2 x. 

tan 2 x = sin 2 x. 
By [12] , tan 2 x = 2 sin x cos x. 

tan 2 x = 2 tan x cos 2 x. 

„ 2 tan x 

tan 2 x = 

sec 2 x 

2 tanx 

1 + tan 2 x" 

tan 2 x + tan 4 x = 2 tan x. 

tan x (tan 3 x + tan x — 2) = 0. 

tan x (tan x — 1) (tan 2 x + tan x + 2) = 0. 



By Prob. 2, Ex. V, tan 2 x = 



TEACHERS' EDITION. 



195 



(i) 
(ii) 

(iii) 



tan x = 0. 
.-. x = 0° or 180°. 
tan x — 1 = 0. 
tan x — 1. 
.-. x = 45° or 225°. 
tan 2 x + tan x 4- 2 = 0. 
4tan 2 x4-() + 1 - - 7. 

2 tan x + 1 = ± V- 7. 

tan x = | ( - 1 ± V- 7). 
.-. x is impossible. 

.-. x = 0°, 45°, 180°, or 225°. 



126. Solve tan x -f cot x = tan 2 x. 

tan x -f cot x = tan 2 x. 
1 2 tan x 

tanx 
tan 2 x + 1 



By [14], 



tan x -f 



1 — tan 2 x 
2 tanx 



tanx 1 - tan 2 x 

1 - tan 4 x = 2 tan 2 x. 
tan 4 x 4- 2 tan 2 x = 1. 
tan 4 x + 2tan 2 x + 1 = 2. 

tan 2 x + 1 = ± V2. 
tan 2 x = - 1 ± V2. 
tan 2 x = 0.4142 or - 2.4142. 
tanx = ±0.6436. 
.-. x = 32°46 / , 147° 14', 212° 46', or 327° 14'. 



-i o« o i 1 - tanx a 

127. Solve = cos 2 x. 



1 + tan x 



By [13], 



1 — tan x 

1 4- tan x 

cos x — sin x 

cos x 4- sin x 

cos x — sin x 



= cos 2 x. 

= cos 2 x — sin 2 x. 

= (cos x — sin x) (cos x 4- sin x). 
cos x 4- sin x 

cos x — sin x = (cos x — sin x) (cos x 4- sin x) 2 . 

(cos x — sin x) [1 — (cos x 4- sin x) 2 ] = 0. 

(i) cos x — sin x = 0. 

cos x = sin x. 

.-. x = 45° or 225°. 



196 PLANE TRIGONOMETRY. 

(ii) 1 — (cos x + sin x) 2 = 0. 

1 — (cos 2 x + sin 2 x + 2 sin x cos x) = 0. 

1 — (1 + 2 sin x cos x) = 0. 

2 sin x cos x = 0. 

sin x cos x = 0. 

.-. x = 0°, 90°, 180°, or 270°. 

.\ x = 0°, 45°, 90°, 180°, 225°, or 270°. 

128. Solve sin x + sin 2 x = 1 — cos 2 x. 

sin x + sin 2 x = 1 — cos 2 x. 
By [12], sin x -f 2 sin x cos x = 1 — cos 2 x. 

By [16], sin x + 2 sin x cos x = 2 sin 2 x. 

sin x (1 + 2 cos x — 2 sin x) = 0. 

(i) sin x = 0. 

.-. x = 0° or!80°. 

(ii) 1 + 2 cos x — 2 sin x = 0. 

sin x — cos x = J. 

sin 2 x -f cos 2 x — 2 sin x cos x = \. 

1 — 2 sin x cos x = J. 

2 sin x cos x = f . 

By [12], sin 2x = { = 0.75. 

.-. 2 x = 48° 36', 131° 24 / , 408° 36', or 491° 24'. 
/. x = 24° 18', 65° 42', 204° 18', or 245° 42'. 
.-. x = 0°, 24° 18', 65° 42', 180°, 204° 18', or 245° 42 7 . 
But the values x = 24° 18 7 and x = 245° 42' do not satisfy the given 
equation. 

.-. x = 0°, 65° 42', 180°, or 204° 18 7 . 

129. Solve sec 2 x + 1 = 2 cos x. 

sec 2 x 4- 1 = 2 cos x. 
+ 1 = 2 cos x. 



cos 2x 

1 -f- cos 2 x = 2 cos x cos 2 x. 
By [13], 1 + cos 2 x — sin 2 x = 2 cos x (cos 2 x — sin 2 x). 

1 -f cos 2 x — 1 + cos 2 x = 2 cos x(cos 2 x — 1 -f cos 2 x). 
2 cos 2 x = 2 cos x (2 cos 2 x — 1). 
cos x (2 cos 2 x — cos x — 1) = 0. 
cos x (cos x — 1) (2 cos x -f 1) = 0. 
(i) cos x = 0. 

.-. x = 90° or 270°. 



teachers' edition. 197 

(ii) cos x — 1 = 0. 

cosx = 1. 

.-. x = 0°. 
(iii) 2 cos x -f 1 = 0. 

cos x = — i. 
.-. x = 120° or 240°. 
.-. x = 0°, 90°, 120°, 240°, or 270°. 

130. Solve tan 2 x + tan 3 x = 0. 

tan 2 x + tan 3 x = 0. 

tan 2 x = — tan 3 x = tan ( — 3 x). 
.-. 2x = -3xor 180° -3x. 
(i) 5x = o + )i360 o . 

.-. x = 0°, 72°, 144°, 216°, or 288°. 

(ii) 5x = 180° + n 360°. 

.-. x = 36°, 108°, 180°, 252°, or 324°. 
.-. x = 0°, 36°, 72°, 108°, 144°, 180°, 216°, 252°, 288°, or 324°. 

131. Solve tan (£ it -f x) -f tan (£ 7r — x) = 4. 

tan (i 7t -f x) + tan (J it — x) =4. 

-« ™-. -. r^^-. 1 + tanx 1 — tanx 

By [6] and [10] , — f- = 4. 

J L J L J 1 - tan x 1 + tan x 

1 -f 2 tan x -f tan 2 x + 1 — 2 tan x + tan 2 x = 4 — 4 tan 2 x. 

6tan 2 x = 2. 

tan 2 x = i. 

tan x = ± Vi = ± i "^8- 

.-. x = 30°, 150°, 210°, or 330°. 

132. Solve Vl + sin x — Vl — sin x = 2 cos x. 



Vl + sin x — Vl — sin x = 2 cos x. 
Square, 

1 + sin x — 2 Vl — sin 2 x + 1 — sin x = 4 cos 2 x. 
2 — 2 Vl - sin 2 x = 4 cos 2 x. 
1 — Vcos 2 x = 2 cos 2 x. 
1 — cosx = 2 cos 2 x. 
2 cos 2 x 4- cos x — 1 = 0. 
(cos x -f 1) (2 cos x — 1) = 0. 
(i) cos x + 1 = 0. 

cos x = — 1. 
.-. x = 180°. 



198 



(ii) 



PLANE TRIGONOMETRY. 



2 cos » — 1=0.. 

cos x = %. 
.-. x = 60° or 300°. 
,-. x = 60°, 180°, or 300°. 
But the values x = 180° and x = 300° do not satisfy the given equation 

/. x = 60°. 



133. Solve tan x tan 3 x = — §. 

tan x tan 3 x = — §. 
By Prob. 19, Ex. XXIV, 

3 tanx — tan 3 x _ 2 
l-3tan 2 x ~~~^ 
3 tan 2 x — tan 4 x 



(i) 



(«) 



tanx- 



l-3tan 2 x D 

15 tan 2 x — 5 tan 4 x = — 2 -f 6 tan 2 x. 
5 tan 4 x - 9 tan 2 x -2 = 0. 
(tan 2 x - 2) (5 tan 2 x + 1) = 0. 

tan 2 x -2 = 0. 
tan 2 x = 2. 

tan x = ± V2 = ± 1.4142. 
.-. x = 54° 44', 125° 16', 234° 44', or 305° 16: 

5tan 2 x + 1 = 0. 
5tan 2 x = — 1. 
tan 2 x = — 1. 



.-. x is impossible. 

.-. x = 54° 44', 125° 16', 234° 44', or 305° 16'. 



134. Solve sin (45° + x) + cos (45° - x) - 1. 

sin (45° + x) + cos (45° - x) = 1. 

Now 45° + x = 90° - (45° - x). 

.-. cos (45° - x) + cos (45° - x) = 1. 

2cos(45°-x) = 1. 

cos (45° - x) = i. 

45° - x = 60° or 300°. 

.-. x = -15°or -255°. 
.-. x = 105° or 345°. 



TEACHEKS* EDITION. 199 

135. Solve tan x 4- sec x — a. 

tan x -f sec x — a. 

sec x = a — tan x. 

sec 2 x = a 2 — 2 a tan x -f tan 2 x. 

By Prob. 2, Ex. V, 1 + tan 2 x = a 2 - 2 a tan x + tan 2 x. 

2 a tan x = a 2 — 1. 

a 2 - 1 

tan x = 

2a 

,a 2 -l 

.-.x = tan -1 

2a 

136. Solve cos 2 x = a (1 — cos x). 

cos 2 x = a (1 — cos x). 
By [13], cos 2 x — sin 2 x = a(l — cosx). 

cos 2 x — 1 4- cos 2 x = a — a cosx. 

2 cos 2 x + a cosx = a + 1. 
16 cos 2 x 4- ( ) + a 2 = a 2 + 8 a + 8. 



4 cosx 4- a = ± Va 2 4- 8 a 4- 8. 

- a ± Va 2 4- 8 a + 8 



, /- a ± Va 2 + 8 a 4- : 
x = cos- 1 ( 



137. Solve (1 — tan x) cos 2 x = a (1 4- tan x). 

(1 — tan x) cos 2 x = a (1 4- tan x). 
1 4- tan x 



cos 2 x = a 
cos 2 x = 
By [13], cos 2 x — sin 2 x = 



1 — tan x 
a (cos x 4- sin x) 
cos x — sin x 

a (cos x 4- sin x) 



cos x — sin x 

(cos x — sin x) 2 (cos x 4- sin x) — a (cos x 4- sin x) = 0. 
(cos x 4- sin x) [(cos x — sin x) 2 — a] = 0. 

(i) cos x 4- sin x = 0. 

sin x — — cos x. 
.-. x = 135° or 315°. 

(ii) (cos x — sin x) 2 — a = 0. 

cos 2 x 4- sin 2 x — 2 sin x cos x = a. 

1 — 2 sin x cos x = a. 



200 PLANE TRIGONOMETRY. 

2 sin x cos x = 1 — a. 
sin 2 x = 1 *- a. 
.-. 2x = sin- 1 (l — a). 
.-. x = isin- 1 (l — a). 
.-. x = 135°, 315°, or i sin- 1 (1 - a). 

138. Solve sin 6 x -f cos 6 x — T 7 2 sin 2 2 x. 

sin 6 x + cos 6 x = T 7 2 sin 2 2 x. 
(sin 2 x -f cos 2 x) (sin 4 x — sin 2 x cos 2 x + cos 4 x) = ^ sin 2 2x. 
sin 4 x — sin 2 x cos 2 x + cos 4 x = ^ sin 2 2x. 
(sin 4 x + 2 sin 2 x cos 2 x + cos 4 x) — 3 sin 2 x cos 2 x = | sin 2 x cos 2 x. 
(sin 2 x -f eos 2 x) 2 = y sin 2 x cos 2 x. 
1 = ia sin 2 x cos 2 . 
16 sin 2 x cos 2 x = 3. 
4 sin 2 x cos 2 x = f. 

sin 2 2x = ± 1V3. 
.-. 2 x = 60°, 120°, 240°, 300°, 420°, 480°, 600°, or 660°. 
.-. x = 30°, 60°, 120°, 150°, 210°, 240°, 300°, or 330°. 

139. Solve cos 3 x + 8 cos 3 x = 0. 

cos 3 x + 8 cos 3 x = 0. 
By Prob. 19, Ex. XIV, 

4 cos 3 x — 3 cos x + 8 cos 3 x = 0. 

12 cos 3 x — 3 cosx = 0. 

cosx (4 cos 2 x — 1) = 0. 

(i) cos x = 0. 

.-. x = 90° or 270°. 
(ii) 4cos 2 x — 1 = 0. 

4 cos 2 x = 1. 
cos 2 x = i. 
cos X = ± J. 
.-. x = 60°, 120°, 240°, or 300°. 
.-. x = 60°, 90°, 120°, 240°, 270°, or 300°. 

140. Solve sec (x + 120°) + sec (x - 120°) = 2 cosx. 

sec (x + 120°) + sec (x - 120°) = 2 cosx. 

-= 2 cosx. 






cos (x + 120°) cos(x - 120°) 

30s (x - 120°) + cos (x + 120°; 

cos (x + 120°) cos (x - 120°) 



cos (x - 120°) + cos (x + 120°) _ 

— -= 2 cosx. 



teachers' edition. 201 

rr%c%n 2 cos x cos 120° 

By T221, = 2 cos x. 

J L J ' cos (x + 120°) cos (x - 120°) 

By [5] and [9], 

2 cos x cos 120° 

= 2 cos x. 

(cos x cos 120° - sin x sin 120°) (cos x cos 120° + sin x sin 120°) 

2 cos x cos 120° 

— 2 cos x 

cos 2 x cos 2 120° - sin 2 x sin 2 120° 

Now sin 120° = \ V3 and cos 120° = - |. 

— cos x 

.-. = 2 cos x. 

£cos 2 x — } sin 2 x 

— cosx = I cos x (cos 2 x — 3 sin 2 x). 

— 2 cosx = cosx (4 cos 2 x — 3). 

cos x (4 cos 2 x — 1) = 0. 

(i) cosx = 0. 

.-. x = 90° or 270°. 

(ii) 4 cos 2 x -1=0. 

4 cos 2 x = 1. 

cos 2 x = i. 

cosx = ± \. 

.-. x = 60°, 120°, 240°, or 300°. 

.-. x = 60°, 90°, 120°, 240°, 270°, or 300°. 

141. Solve esc x = cot x -f Vs. 

esc x = cot x + V3. 
csc 2 x = cot 2 x + 2 V3 cot x + 3. 
By Prob. 3, Ex. V, 1 + cot 2 x = cot 2 x + 2 V3 cot x + 3. 
2 V3 cot x = - 2. 

cot x = —~t V§. 

V3 
.-. x = 120° or 300°. 
But the value x = 300° does not satisfy the given equation. 

.-. x = 120°. 

142. Solve 4 cos 2 x + 6 sin x = 5/ 

4 cos 2 x + 6 sin x = 5. 

By [13], 4 (1 - 2 sin 2 x) + 6 sin x = 5. 

4 — 8 sin 2 x + 6 sin x = 5. 

8sin 2 x - 6sinx + 1 = 0. 

(2 sin x - 1) (4 sin x - 1) = 0. 



202 PLANE TRIGONOMETRY. 

(i) 2 sin x - 1 = 0. 

2sinx = 1. 
sin x — \. 
.•: x = 30° or 150°. 
(ii) 4 sin x — 1 = 0. 

4 sin x = 1. 
sin x = £ = 0.25. 
.-. x = 14° 29' or 165° 81'. 
.-. x = 14° 29', 30°, 150°, or 165° 81', 

143. Solve cos x — cos 2 x = 1. 

cos x — cos 2 x = 1. 

By [13], cos x — cos 2 x + sin 2 x = 1. 

cos x — cos 2 x + 1 — cos 2 x = 1. 

2 cos 2 x — cosx = 0. 

cos x (2 cos x — 1) = 0. 

(i) cos x = 0. 

... x = 90° or 270°. 
(ii) 2 cos x — 1 ' = 0. 

2 cosx = 1. 

COS X = -J-. 

.-. x = 60° or 300°. 

.-. x = 60°, 90°, 270°, or 300°. 

144. Solve sin 4 x — sin 2 x = sin x. 

sin 4 x — sin 2 x = sin x. 
By [21], 2 cos 3 x sin x = sin x. 

sin x (2 cos 3 x — 1) = 0. 
(i) sin x = 0. 

.-. x = 0° or 180°. 
(ii) 2cos3x-l = 0. 

2 cos3x = 1. 
cos 3 x = i. 
.-. 3 x = 60° + n 360° or 300° + n 360°. 

, x = 20°, 140°, or 260° ; or 100°, 220°, or 340°. 



. x = 20°, 140°, or 260° ; or 100°, 220 u , or Mi) u . 
. x = 0°, 20°, 100°, '140°, 180°, 220°, 260°, or 340°. 



145. Solve 2 sin 2 x + sin 2 2 x = 2. 

2sin 2 x + sin 2 2x = 2. 

By [12], 2 sin 2 x + 4 sin 2 x cos 2 x = 2. 

sin 2 x + 2 sin 2 x (1 — sin 2 x) = 1. 



TEACHERS 7 EDITION. 203 

sin 2 x + 2 sin 2 x — 2 sin 4 x = 1. 

2 sin 4 x — 3 sin 2 x +1 = 0. 

(sin 2 x- l)(2sin 2 x - 1) = 0. 

(i) sin 2 x -1 = 0. 

sin 2 x = 1. 
sin x = ± 1. 
.-. x = 0°, 90°, 180°, or 270°. 

(ii) 2 sin 2 x -1 = 0. 

2sin 2 x = 1. 

sin 2 x = £. 

sin x = ± V| — ± i V2. 
.-. x = 45°, 135°, 225°, or 315°. 
.-. x = 0°, 45°, 90°, 135°, 180°, 225°, 270°, or 315°. 
But the values x = 0° and x = 180° do not satisfy the given equation. 
.-. x = 45°, 90°, 135°, 225°, 270°, or 315°. 

146. Solve cos 5 x 4- cos 3 x + cos x = 0. 

cos 5x + cos 3x4- cos x = 0. 

By [22], 2 cos4xcosx + cosx = 0. 

cos x (2 cos 4 x 4- 1) = 0. 

(i) cos x = 0. 

.-. x = 90° or 270°. 

(ii) 2 cos 4 x 4- 1 = 0. 

2 cos4x = — 1. 
cos4x = — -J-. 
.-. 4 x = 120° 4- n 360° or 240° + n 360°. 
.-. x = 30°, 120°, 210°, or 300° ; or 60°, 150°, 240°, or 330°. 
.-. x = 30°, 60°, 90°, 120°, 150°, 210°, 240°, 270°, 300°, or 330°. 

147. Solve sec x — cot x = esc x — tan x. 

sec x — cot x = esc x — tan x. 
1 cos x 1 sin x 



cos x sin x sin x cos x 
sin x — cos 2 x = cos x — sin 2 x. 

(sin x — cos x) + (sin 2 x — cos 2 x) = 0. 

(sin x — cos x) (1 + sin x 4- cos x) = 0. 

(i) sin x — cos x = 0. 

sin x = cos x. 
.-. x = 45° or 225°. 



204 PLANE TRIGONOMETRY. 

(ii) 1 + sin x + cos x = 0. 

since + cosx =— 1. 
sin 2 x -f 2 sin x cos x -f cos 2 x = 1. 
1+2 since cosx = 1. 
2 sin x cos x = 0. 
By [12], sin2x = 0. . 

.-, 2x = 0°, 180°, 360°, or 540°. 
.-. x = 0°, 90°, 180°, or 270°. 
.-. x = 0°, 45°, 90°, 180°, 225°, or 270°. 

148. Solve tan 2 x + cot 2 x = *£-. 

tan 2 x + cot 2 x = \ -. 

1 10 

tan 2 x H = — 

tan 2 x 3 

3tan 4 x + 3 = 10 tan 2 x. 
3 tan*x - 10 tan 2 x + 3 = 0. 
(tan 2 x - 3) (3 tan 2 x - 1) = 0. 
(i) tan 2 x-3 = 0. 

tan 2 x = 3. 
tan x = ± V3. 
.-. x = 60°, 120°, 240°, or 300°. 
(ii) 3tan 2 x-l=0. 

3tan 2 x = 1. 
tan 2 x = -J-. 

tan x = ± V^ — ± % Vs. 
.-. x = 30°, 150°, 210°, or 330°. 
.-. x = 30°, 60°, 120°, 150°, 210°, 240°, 300°, or 330°. 

149. Solve sin 4 x — cos 3 x = sin 2 x. 

sin 4 x — cos 3 x = sin 2 x. 

sin 4 x — sin 2 x — cos 3 x = 0. 

By [21], 2 cos 3 x sin x — cos3x = 0. 

cos 3 x (2 sin x — 1) = 0. 

(i) cos 3 x = 0. 

.-. 3 x = 90° + n 360° or 270° + n 360°. 
.-. x = 30°, 150°, or 270° ; or 90°, 210°, or 330°. 
(ii) 2 sin x - 1 = 0. 

2sinx = 1. 
sin x = i. 
.-. x = 30° or 150°. 
... x = 30°, 90°, 150°, 210°, 270°, or 330°. 



teachers' edition. 205 

150. Solve sin x + cos x = sec x. 

sin x + cos x = sec x. 

1 

sin x + cos x = . 

cosx 

sin x cos x 4- cos 2 x = 1. 

sin x*cos x + (cos 2 x — 1) = 0. 

sin x cos x — sin 2 x = 0. 

sin x (cos x — sin x) = 0. 

(i) sin x = 0. 

.-. x = 0° or 180°. 

(ii) cos x — sin x = 0. 

cos x = sin x. 

.-. x = 45° or 225°. 

.-. x = 0°, 45°, 180°, or 225°. 

151. Solve 2 cos x cos 3 x + 1 =0. 

2 cos x cos 3 x + 1 = 0. 
By Prob. 19, Ex. XIV, 

2 cos x (4 cos 3 x — 3 cos x) -f 1 = 0. 

8 cos 4 x - cos 2 x + 1 = 0. 

(4 cos 2 x -1) (2 cos 2 x - 1) = 0. 

(i) 4cos 2 x-l = 0. 

4 cos 2 x = 1. 

cos 2 x = \. 

COS X = ± i 

.-. x = 60°, 120°, 240°, or 300°. 

(ii) 2cos 2 x-l = 0. 

2 cos 2 x = 1. 

cos' 2 x = \. 

cos X = ± Vj = ± I V2. 
.-. x = 45°, 135°, 225°, or 315°. 
.-. x = 45°, G0° ? 120°, 135°, 225°, 240°, 300°, or 315°. 

152. Solve cos 3x — 2 cos 2x 4- cosx =0. 

cos 3 x — 2 cos 2 x -f cos x = 0. 

(cos 3 x + cos x) — 2 cos 2 x = 0. 

By [22], 2cos2xcosx - 2cos2x = 0. 

cos 2 x (cosx — 1) = 0. 

(i) cos 2 x = 0. 

.-. 2x = 90°, 270°, 450°, or 630°. 
.-. x = 45°, 135°, 225°, or 315°. 



206 PLANE TRIGONOMETRY. 

(ii) cos x — 1 = 0. 

cosx = 1. 
.-. x = 0°. 
.-. x = 0°, 45°, 135°, 225°, or 315°. 

Solve tan 2 x tan x == 1. 

tan 2 x tan x = 1. 
tan 2 x : 



tanx 
tan 2 x = cot x. 
.-. 2x = 90°-xor270°-x. 

3 x = 90° + n 360° or 270° + n 360°. 
, x = 30°, 150°, or 270° ; or 90°, 210°, or 330°. 
, x = 30°, 90°, 150°, 210°, 270°, or 330°. 



154. Solve sin (x + 12°) +, sin (x - 8°) = sin 20°. 

sin (x + 12°) + sin (x - 8°) = sin 20°. 
By [20], 2 sin (x + 2°) cos 10° = sin 20°. 

2 sin (x + 2°) cos 10° = 2 sin 10° cos 10°. 
2 sin (x + 2°) = 2 sin 10°. 
sin (x + 2°) = sin 10°. 
.-. x + 2° = 10° or 170°. 
.-. x = 8° or 168°. 



155. Solve tan (60° + x) tan (60° - x) = - 2. 

tan (60° + x) tan (60° - x) = - 2. 
By [6] and [10], 

tan 60° + tan x tan 60° - tan x 

x = - 2. 



1 — tan 60° tan x 1 -f tan 60° tan x 
tan 2 60°- tan 2 x 
1 -tan 2 60° tan 2 x 
3-tan 2 x 



= -2. 
= -2. 



1 -3tan 2 x 
3 - tan 2 x --2 + 6 tan 2 x. 
7tan 2 x = 5. 
tan 2 x = f = 0.71428571. 
tanx =±0.8451. 
.-. x = 40° 12', 139° 48 7 , 220° 12 7 , or 319° 48'. 



teachers' edition. ■ 207 



156. Solve sin (x + 120°) + sin (x + 60°) = $ 

sin (x + 120°) + sin (x + 60°) = f . 

By [20], 2 sin (x + 90°) cos 30° = §. 

2 cos x cos 30° = f . 

2 cos x (i V3) = f . 

3 



2V3 
.-. x = 30° or 330°. 



157. Solve sin (x + 30°) sin (x - 30°) = J. 

sin (x + 30°) sin (x - 30°) = -J-. 
By [23], - i (cos 2 x - cos 60°) = -£-. 

cos 2 x — cos 60° = — 1. 
cos2x — i = — 1. 
cos 2 x = — £. 
.-. 2 x = 120°, 240°, 480°, or 600°. 
.-. x = 60°, 120°, 240°, or 300°. 

158. Solve sin 4 x -f cos 4 x = f . 

sin 4 x + cos 4 x = f. 
sin 4 x + 2 sin 2 x cos 2 x 4- cos 4 x = 2 sin 2 x cos 2 x -f |. 
(sin 2 x + cos 2 x) 2 = 2 sin 2 x cos 2 x + |. 
1 = 2sin 2 xcos 2 x + f. 
2 sin 2 x cos 2 x = |. 
4 sin 2 x cos 2 x = f. 
By [12], shi 2 2x = |. 

sin 2x = ±l-Vs. 
.-. 2x = 60°, 120°,, 240°, 300°, 420<\ 480°, 600°, or 660°. 
.-. x = 30°, 60°, 120°, 150°, 210°, 240°, 300°, or 330°. 

159. Solve sin 4 x — cos 4 x = ^. 

sin 4 x — cos 4 x = ¥ V 
(sin 2 x -f cos 2 x) (sin 2 x — cos 2 x) = £ z . 
sin 2 x — cos 2 x = 2^. 
cos 2 x — sin 2 x = — 2 7 5 . 
By [13], cos2x = - ^V 

.-. 2 x = 106° 16', 253° 44', 466° 16 / , or 613° 44'. 
.-. x = 53° 8', 126° 52', 233° 8', or 306° 52', 



208 - PLANE TRIGONOMETRY. 

160. Solve tan (x 4 30°) - 2 cos x. 

tan (x + 30°) = 2 cos x. 

Let x 4 30° = y. 

Then x = y - 30°. 

Substitute, tan y = 2 cos (?/ — 30°). 

By [9], tan y = 2 cos 2/ cos 30° 4 2 sin 2/ sin 30°. 

sin y rz , . 

= v 3 cos ?/ 4- sm ?/. 

cos 2/ 

sin y = V3 cos 2 ?/ 4 sin y cos ?/. 

sin y (1 — cos 2/) = V3 cos 2 ?/. 

sin 2 y (1 — cos ?/) 2 = 3 cos 4 ?/. 

(1 — cos 2 ?/) (1 — 2 cos?/ 4 cos 2 ?/) = 3 cos 4 ?/. 

1 — 2 cos y 4 2 cos 3 ?/ — cos 4 ?/ = 3 cos 4 ?/. 

4 cos 4 y — 2 cos 3 ?/ 4 2 cos ?/ — 1 = 0. 

(2 cos ?/ - 1) (2 cos 3 ?/ 4 1) = 0. 

(i) 2 cos?/ -1 = 0. 

2 cos?/ = 1. 
cos?/ = 1. 
.-. ?/ = 60° or 300°. 
.-. x = 30° or 270°. 

(ii) 2 cos 3 ?/ 4 1 = 0. 

2 cos 3 ?/ =— 1. 
cos 3 ?/ =— |. 

cos?/ = - Vi. 
The two other roots of equation (ii) are complex. 

cos?/ = -0.7937. 
.-. y = 127° 28 r or 232° 32 7 . 
.-. x = 97° 28' or 202° 32'. 
.-. x = 30°, 97° 28', 202° 32', or 270°. 
But the values x = 97° 28', x = 202° 32', and x = 270° do not satisfy 
the given equation. 

.-. x = 30°. 

161. Solve sec x = 2 tan x 4 \. 

sec x = 2 tan x 4 \. 
sec 2 x = 4 tan 2 x 4 tan x 4 ^g. 
By Prob. 2, Ex. V, 1 4 tan 2 x = 4 tan 2 x 4 tan x 4 T V 
3tan 2 x4 tanx -\% - 0. 
48 tan 2 x 4 16 tan x - 15 = 0. 
(4 tan x 4- 3) (12 tan x - 5) = 0. 



209 

(i) 4 tan x + 3 = 0. 

4 tan x — — 3. 
tanx = - J = -0.75. 
.-. x = 143° 8' or 323° 8'. 

(ii) 12 tan x — 5 = 0. 

12 tan x = 5. 
5 
tanx = — = 0.4167. 
12 

.-. x = 22° 37' or 202° 37'. 

.-. x = 22° 37', 143° 8', 202° 37', or 323° 8'. 

But the values x = 202° 37' and x = 323° 8' do not satisfy the given 

equation. 

.\ x = 22° 37' or 143° 8'. 

162. Solve sin 11 x sin 4 x -f sin 5 x sin 2 x = 0. 

sin 11 x sin 4 x + sin 5 x sin 2 x = 0. 
By [23], sin 11 x sin 4 x = — £ (cos 15 x — cos 7 x), 

and sin 5 x sin 2 x = — \ (cos 7 x — cos 3 x). 

•'• — i ( c °s 15 x — cos 7 x) — i (cos 7 x — cos 3 x) = 0. 
cos 15 x — cos 7 x + cos 7 x — cos 3 x = 0. 
cos 15 x — cos 3 x = 0. 
By [23], - 2 sin i (15 x + 3 x) sin i (15 x - 3 x) = 0. 

— sin 9 x sin 6 x = 0. 
sin 9 x sin 6 x = 0. 
By Prob. 18, Ex. XIV, sin 9 x = 3 sin 3 x - 4 sin3 3 x, 



and by [16], sin 6 x = ± a/— 



cos 3 x 



Substitute, (3 sin 3 x - 4 sin3 3 x) ( ± -J- !^!i?) = o. 

(sin3x)(3-4sin23x)(i: J 1 ~ COS3X ) = 0. 

(i) sin 3 x = 0. 

.-. 3 x = 0°, 180°, 360°, 540°, 720°, or 900°. 
.-. x = 0°, 60°, 120°, 180°, 240°, or 300°. 

(ii) 3-4sin 2 3x = 0. 

4sin23x = 3. 

2 sin 3 x = ± V3. 
sin 3 x = ± i Vs. 
.'. 3 x = 60°, 120°, 240°, 300°, 420°, 480°, 600°, 660°, 780°, 840°, 960°, 
or 1020°. 



210 PLANE TRIGONOMETRY. 

.\x = 20°, 40°, 80°, 100°, 140°, 160°, 200°, 220°, 260°, 280°, 320°, 
or 340°. 
..... /l — cos 3 x 



2 
1 — cos 3 x 



0. 



2 

1 — cos 3 x = 0. 
cos 3 x = 1. 
.-. 3x = 0°, 360°, or 720°. 
.-. x = 0°, 120°, or 240°. 
.-. x = 0°, 20°, 40°, 60°, 80°, 100°, 120°, 140°, 160°, 180°, 200°, 220°, 
240°, 260°, 280°, 300°, 320°, or 340°. 

163. Solve cos x + cos 3 x + cos 5 x + cos 7 x = 0. 

cos x + cos 3 x -f cos 5 x -f cos 7 x = 0. 

(cos7x -f cos5x) -f (cos3x-f cosx) = 0. 

By [22], 2 cos 6 x cos x + 2 cos 2 x cos x = 0. 

cos x (cos 6x + cos 2 x) = 0. 

By [22], cosx x 2 cos 4 x cos 2 x = 0. 

cos x cos 2 x cos 4 x = 0. 

(i) cos x = 0. 

.-. x = 90° or 270°. 

(ii) cos 2 x = 0. 

.-. 2 x = 90°, 270°, 450°, or 630°. 
.-. x = 45°, 135°, 225°, or 315°. 

(iii) cos 4 x = 0. 

.-. 4x = 90°, 270°, 450°, 630°, 810°, 990°, 1170°, or 1350°. 

.-. x = 22i°, 67£°, 112£°, 157*°, 202i°, 247i°, 292£°, or 337*°. 

.-. x = 22£°, 45°, 67i°, 90°, 112}°, 135°, 157*°, 202-J , 225°, 247i°, 270°, 
292i°, 315°, or 337i°. 

164. Solve sin (x + 12°) cos (x - 12°) = cos 33° sin 57°. 

sin (x -1- 12°) cos (x - 12°) = cos 33° sin 57°. 
By [20] , i (sin 2 x + sin 24°) = -J- (sin 90° + sin 24°). 

sin 2 x + sin 24 = sin 90° + sin 24°. 
sin 2 x = sin 90°. 
.-. 2&=.90° + ft 360°. 
.*. x = 45° or 225°. 



teachers' edition. 211 

165. Solve sin - 1 x + sin - 1 i x = 120°. 

sin- 1 ^ + sin-Hz = 120°. 
sin(sin- 1 x + sin-H^) = sin 120° = cos 30° = -J-V3. 
By [4], sin (sin - x x) cos (sin — 1 i x) + cos (sin -!£) sin (sin- 1 -J- £) = -§- V3. 

xVl -±x 2 + ix Vl -x 2 = \ Vs. 
x V4-x 2 +x Vl-x 2 = Vs. 

x V4 -x 2 = V3 - x Vl - x 2 . 
x 2 (4 - x 2 ) = 3 - 2 V3xVl -x 2 + x 2 (l - x 2 ). 
4x 2 -x 4 = 3-2V3x Vl -x 2 + x 2 - x 4 . 
3x 2 -3 = -2 V3x Vl -x 2 . 
Square, 9 x 4 - 18 x 2 + 9 = 12 x 2 - 12 x 4 . 

21 x 4 - 30 x 2 + 9 = 0. 
7 x 4 - 10 x 2 -f 3 = 0. 
(x 2 - 1) (7 x 2 - 3) = 0. 

(i) x 2 - 1 = 0. 

x 2 = l. 
.-. x= ± 1. 

(ii) 7 x 2 - 3 = 0. 

7 x 2 = 3. 
x 2 = f 
.-. x = ± V| = ± } V21. 
.-. x = ± 1 or ± I V2I. 



166. Solve tan- 1 x + tan- 1 2 x = tan- x 3 V3. 

tan- 1 x + tan- 1 2 x = tan- 1 3 V3. 
tan (tan- 1 x + tan- * 2 x) = 3 V3. 

By [6], ^-±1^ = 3V3. 



-^-.3V3. 
1 - 2 x 2 

= V3. 



1 -2x 2 

x = V3-2 V3x 2 . 
2V3x 2 + x-V3 = 0. 
x 2_|_iV3x-|- = 0. 
{x - i VS) (x + i Vs) = 0. 

.-. x = iV3 or-iV3. 



212 PLANE TRIGONOMETRY. 

167. Solve sin- 1 x + 2 cos- ] x = f it. 

sin- 1 x -h 2cos _1 x = § it. 
Now sin-ix -f cos -1 x = \it. 

Subtract, cos- x x = £ ^ 



X : 



Vs. 



168. Solve sin- 1 x + 3 cos- 1 x = 2 10°. 

sin-ix + 3 cos-!x = 210°. 
But sin- 1 x + cos- x x = 90°. 

Subtract, 2 cos~ * x = 120°. 

cos-ix^O . 
.-.x = i. 

169. Solve tan-!x + 2 cot-ix = 135°. 

tan-ix + 2 cot- x x = 135°. 
But tan-!x + coWx = 90°. 

Subtract, cot~ 1 x = 45°. 

.-. x = 1. 

170. Solve tan^ 1 (x -f 1) + tan- 1 (x - 1) = tan- 1 2 x. 

tan- 1 (x + 1) + tan- 1 (x - 1) = tan- 1 2 x. 
tan [tan- 1 (x + 1) + tan- 1 (x - 1)] = 2 x. 
By [6], x+l+x-1 =2x> 

1 -(x+l)(x- 1) 

■ = 2x. 



2-x 2 

x = 2 x — x 3 . 
x 3 - x = 0. 
x(x 2 -l) = 0. 

.-. x = or ± 1. 

171. Solve tan- 1 5^ + tan- 1 — | = ? tt. 
x + 1 x - 1 4 

-x + 2 , . .x-2 3 

tan— 1 h tan -1 = -7t. 

x+ 1 x - 1 4 

r~ />• _J_ 9 x 2~I 

tan tan -1 - — : \- tan -1 = tan f it = — 1. 

L x + 1 x-lJ 

x + 2 x-2 

~ ™ X -f 1 X — 1 ^ 

1 X 

X + 1 X — 1 



teachers' edition. 213 

( X _ i) ( X + 2) + (x + 1) (x - 2) = 1 
(x + 1) (x - 1) 



(a; 


+ : 


2)(x- 


-2) 








2x 2 • 
3 


-4 


:- 1. 






2x 2 - 
i 


-4 = 

2x 2 = 

x 2 = 

\ X = 


-3. 

1. 
h 



172. Solve tan- 1 -^- = 60°. 
1 -x 2 

tan-i- 2 ^- =60°. 
1 -x 2 

By [14], 2tan- 1 x = 60°. 

tan- 1 x = 30°or 210°. 

.-. x = tan 30° or tan 210°. 

r.X = i VS. 



173. Solve cos 2 sec + sec + 1 = 0. 

By [13], 



cos 2 sec + sec + 1 


= 0. 


cos 2 - sin 2 

1 qpr» /? 1.1 


= 0. 


cos 


cos 2 0-sin 2 1 

n + ^ +1 

cos cos 


= 0. 


2cos 2 0- 1 1 

5 + n+ 1 

cos cos 


= 0. 


2cos 2 - 1 + 1 + cos0: 


= 0. 


2 COS 2 + COS : 


= 0. 


cos (2 cos + 1) 


= 0. 


cos 


= 0. 


.-. 


= 90° or 270°. 



(i) 



(ii) 2 cos + 1 = 0. 

2 cos = — 1. 
cos = — J. 
.-. = 120° or 240°. 
.-. = 90°, 120°, 240°, or 270°. 
But the values = 90° and = 270° do not satisfy the given equation. 

.-. = 120° or 240°. 



214 PLANE TRIGONOMETRY. 

174. Solve sin x cos 2 x tan x cot 2 x sec x esc 2 x = 1. 

sin x cos 2 x tan x cot 2 x sec x esc 2 x = 1. 

By [13], cos 2 x = cos 2 x — sin 2 x. 

cot 2 x — 1 
By [15], cot2x = 



2 cotx 
cos 2 x 
sin 2 x cos 2 x — sin 2 x 



cos x 2 sin x cos x 

sin x 
1 1 



By [3] and [13], csc2x: 

sin 2 x 2 sin x cos x 

Substitute, 

, „ .. ■ . /sin x\ /cos 2 x — sin 2 x\ / 1 \/ 1 \ 

sin x (cos 2 x - sin 2 x) ( ) ( ) ( ) ( ) = i . 

Vcosx/ \ 2 sin x cos x / Vcosx/ \2 sin x cos x/ 

(cos 2 x — sin 2 x) 2 



4 cos 4 x 
(2cos 2 x- l) 2 
4cos 4 x 
2 cos 2 x — 1 



= 1. 

= 1. 



2 cos 2 x 
.-. 2 cos 2 x — 1 = ± 2 cos 2 x. 
4cos 2 x = 1. 
cos 2 x = J. 
cos x = ± \. 
.-. x = 60°, 120°, 240°, or 300°. 

175. Solve sin ix (cos 2 x - 2) (1 - tan 2 x) = 0. 
sin i x (cos 2 x — 2) (1 — tan 2 x) = 0. 
(i) sin£x = 0. 

.-. ix = 0°or 180°. 
.-. x = 0°. 
(ii) cos 2 x - 2 = 0. 

cos 2 x = 2. 
.*. x is impossible, 
(iii) 1 - tan 2 x = 0. 

tan 2 x = 1. 
tan x = ± 1. 
.-. x = 45°, 135°, 225°, or 315°. 
... x = 0°, 45°, 135°, 225°, or 315°. 



215 



176. Solve sin 3 x = cos 2 x - 1. 

sin 3 x = cos 2 x — 1. 
By Prob. 18, Ex. XIV, and [13], 

3 sinx — 4 sin 3 x = 1 — 2 sin 2 x — 1. 
4 sin 3 x — 2 sin 2 x — 3 sin x = 0. 
sin x (4 sin 2 x - 2 sin x — 3) = 0. 
(i) sin x = 0. 

■\ x = 0° or 180°. 
(ii) 4 sin 2 x - 2 sin x - 3 = 0. 

16sin 2 x -8sinx + 1 = 13. _ 
4 sin x - 1 = ± Vl3. 
4 sin x = 1 i Vl3. 
sinx = i(l± Vl3). 
sinx = 1.1514 or - 0.6514. 
.-. x = 220° 39' or 319° 21'. 
.-. x = 0°, 180°, 220° 39', or 319° 2V. 

177. Solve tan x -f tan 2 x = 0. 

tan x + tan 2 x = 0. 

By [14], tanx + i 2t f : ! ■ 0. 

1 — tan 2 x 

tan x — tan 3 x + 2 tan x = 0. 

tan x (tan 2 x - 3) = 0. 

(i) tan x = 0. 

.-. x = 0° or 180°. 

(ii) tan 2 x -3 = 0. 

tan 2 x = 3. 

tan x = ± V3. 

.-. x = 60°, 120°, 240°, or 300°. 

... x = 0°, 60°, 120°, 180°, 240°, or 300° 

178. Solve sin 2 6 = cos 3 0. 

sin 2 = cos 3 6. 
By [12], and Prob. 19, Ex. XIV, 

2 sin 6 cos = 4 cos 3 — 3 cos 0. 
4 cos 3 - 3 cos 6 - 2 sin cos 6 = 0. 
cos0(4cos 2 0-3 - 2 sin 6) = 0. 

!(i) cos = 0. 

.-. = 90° or 270°. 



216 PLANE TRIGONOMETRY. 

(ii) 4cos 2 0-3-2sin0 = O. 

4-4 sin 2 - 3 - 2 sin = 0. 

4sin 2 + 2sin0- 1. 

16sin 2 0-f 8 sin 6 + 1 = 5. 

4 sin 4- 1 = ± V5. 
4 sin = - 1 ± Vs. 
sin0 = \{- 1 ±V5) 

= H-1 ±2.23606) 
= 0.3090 or -0.8090. 
.-. = 18° or 162° ; or 234° or 306°. 
.-. = 18°, 90°, 162°, 234°, 270°, or 306°. 

179. Solve (3-4 cos 2 x) sin 2 x = 0. 

(3 -4cos 2 x)sin2x = 0. 

(i) 3-4cos 2 x = 0. 

4 cos 2 x = 3. 
2cosx = ± V3._ 
' cos x=^zi V3. 

.♦. x = 30°, 150°, 210°, or 330°. 

(ii) sin 2 x = 0. 

.-. 205 = 0°, 180°, 360°, or 540° 
.-. x = 0°, 90°, 180°, or 270°. 
.♦. x = 0°, 30°, 90°, 150°, 180°, 210°, 270°, or 330°. 

180. Solve sin x 4 sin 2 x 4- sin 3 x = 0. 

sin x 4- sin 2 x 4 sin 3 x = 0. 

(sin 3x4 sin x) 4 sin 2 x = 0. 

By [20], 2 sin 2 x cos x 4 sin 2 x = 0. 

sin 2 x (2 cos x 4- 1) = 0. 

(i) sin 2 x = 0. 

.-. 2x = 0°, 180°, 360°, or 540°. 
.% x = 0°, 90°, 180°, or 270°. 

(ii) 2 cos x 4 1 = 0. 

2cosx =— -1. 
cos x = — "-J-. 
.-. x = 60°, 120°, 240°, or 300°. 
.-. x = 0°, 60°, 90°, 120°, 180°, 240°, 270°, or 300°. 
But the values x = 60° and x = 300° do not satisfy the given equation. 
.-. x = 0°, 90°, 120°, 180°, 240°, or 270°. 



teachers' edition. 217 

181. Solve sin + 2 sin 2 + 3 sin 3 = 0. 

sin (9 + 2 sin 2 + 3 sin 3 = 0. 
By [13], and Prob. 18, Ex. XIV, 

sin + 4 sin cos + 9 sin - 12 sin 3 = 0. 

10 sin + 4 sin cos - 12 sin 3 (9 = 0. 

sin0(5 + 2cos0 - 6sin 2 0) = 0. 

(i) sin B = 0. 

.-. = 0° or 180°. 

(ii) 5 + 2 cos - 6 sin 2 = 0. 

5 + 2 cos 0-6 + 6 cos 2 = 0. 

6cos 2 + 2cos0= 1. 

36cos 2 + 12 cos + 1 = 7. 

6 cos + 1 = ± V7. 
6 cos = - 1 ± V7. 
cos0 = i(- 1 ± V7) 

= i(-l ±2.64575) 
= 0.2743 or - 0.6076. 
... 6 = 74° 5 7 or 285° 55 y ; or 127° 25 x or 232° 35 7 . 
.-. = 0°, 74° 5', 127° 25', 180°, 232° 35 7 , or 285° 55 7 . 

182. Solve sin 2 x cos 2 x — cos 2 x — sin 2 cc + 1 = 0. 

sin 2 x cos 2 x — cos 2 x — sin 2 x + 1 = 0. 

sin 2 x cos 2 x — (cos 2 x + sin 2 x) +1=0. 

sin 2 x cos 2 x — 1 + 1 = 0. 

sin 2 x cos 2 x = 0. 

sin x cos x = 0. 

2 sin x cos x = 0. 

sin 2 x = 0. 

.-. 2x = 0°, 180°, 360°, or 540°. 

.-. x = 0°, 90°, 180°, or 270°. 

183. Solve sin x + sin 3 x = cos x — cos 3 x. 

sin x + sin 3 x = cos x — cos 3 x. 
By Probs. 18 and 19, Ex. XIV, 

sin x + 3 sin x — 4 sin 3 x = cos x — 4 cos 3 x + 3 cos x. 
4 sin 3 x — 4 cos 3 x — 4 sin x + 4 cos x = 0. 
(sin 3 x — cos 3 x) — (sin x — cos x) = 0. 
(sin x — cos x) (sin 2 x + sin x cos x + cos 2 x) — (sin x — cos x) = 0. 
(sin x — cos x) (1 + sin x cos x) — (sin x — cos x) = 0. 
(sin x — cos x) (1 + sin x cos x — 1) = 0. 
(sin x — cos x) (sin x cos x) = 0. 



218 



PLAKE TRIGONOMETRY. 



(i) 

(ii) 
(iii) 



sin x — cos x = 0, 

sin x = cos x. 

.-. x = 45° or 225°. 

sin x = 0. 
.-. x = 0° or 180°. 

cos x = 0. 
,\ x = 90° or 270°. 

.♦. x = 0°, 45°, 90°, 180°, 225°, or 270°. 



184. Solve (1 — Vl — tan 2 x) cos 2 x vers 3 x -. 
(1- VT 

(i) 



(ii) 



(iii) 



tan 2 x) cos 2 x vers 3 x = 0. 
1 - Vl - tan 2 x = 0. 



1 = Vl - tan 2 x. 

1 = 1- tan 2 x. 
tan 2 x = 0. 
tan x = 0. 
.-. x = 0° or 180°. 

cos 2 x = 0. 
,\ 2x = 90°, 270°, 450°, or 630°. 
.-. x = 45°, 135°, 225°, or 315°. 

vers 3 x = 0. 
1 — cos 3 x = 0. 
cos3x == 1. 
.-. 3x = 0°, 360°, or 720°. 
.-. x = 0°, 120°, or 240°. 
/. x = 0°, 45°, 120°, 135°, 180°, 225°, 240°, or 315°. 



185. Solve tan (6 + 45°) = 8 tan 0. 

tan (0 + 45°) = 8 tan-0. 

_ _._ tan 6 + tan 45° 

By [6], — — — 8 tan 0. 



1 — tan tan 45° 
tan + 1 



= 8 tan 0. 



1 - tan 
tan + 1 = 8 tan - 8 tan 2 0. 
8 tan 2 0-7tan0=-l. 
256tan 2 0-() + 49 = 17. _ 
16 tan - 7 = ± Vl7. 
16 tan = 7 ± Vl7. 



teachers' edition. 219 

tan e = T \ (7 ± Vrf) 
= A(7± 4.1231) 

= 0.6952 or 0.1798. 
.-. = 34° 48' or 214° 48' ; or 10° 12' or 190° 12'. 
.-. = 10° 12', 34° 48', 190° 12', or 214° 48'. 

186. Solve sin (x - 30°) = f Vs sin x. 

sin (x - 30°) = i V3 sin x. 
By [8] , sin x cos 30° — cos x sin 30° = i V.3 sin x. 
£ V3 sin x — £ cos x = i V3 sin x. 
— i cos x = 0. 
cos x = 0. 
.-. x = 90° or 270°. 

187. Solve tan (0 + 45°) tan = 2. 

tan (0 + 45°) tan = 2. 

t _ __ tan + tan 45° „ _ 

By [6], x tan = 2. 

L J 1 -tan tan 45° 

(tan0 + l)tan0_ 



1 - tan 

tan 2 + tan 



2. 



1 - tan 
tan 2 + tan = 2 - 2 tan 0. 

tan 2 + 3tan0 = 2. 
4tan 2 + () + 9 = 17. 

2 tan + 3 = ± Vl7. 

2 tan = - 3 ± Vl7. 
tan0 = |(-3 ±Vl7) 
= i(- 3 ±4.1231) 
= 0.5616 or -3.5616. 
.-. = 29° 19' or 209° 19' ; or 105° 41' or 285° 41'. 
.-. = 29° 19', 105° 41', 209° 19', or 285° 41'. 

188. Solve sin- 1 ix = 30°. 

sin~H^ = 30°. 
sin (sin— l -J x) = sin 30°. 
ix = i. 
.-. x = 1. 

189. Solve for x and y the system 

x sin a -f 2/ sin p = a, (1) 

x cos a + y cos /3 = 6. (2) 



220 PLANE TRIGONOMETRY. 

Multiply (1) by cos or, x sin a cos a -f y sin /3 cos a = a cos a. (3) 

Multiply (2) by sin a, x sin a cos a + y cos sin a = 6 sin a. (4) 

Subtract (4) from (3), y (sin j3 cos a — cos /3 sin a) — a cos a: — b sin a. 

By [8], y sin (0 — a) = a cos a — b sin a. 

a cos a — b sin a 

•'► 2/ = 

sin (/3 — a) 

Multiply (1) by cos /3, x sin a cos /3 -f y sin /3 cos /3 = a cos /3. (5) 

Multiply (2) by sin /3, x cos a sin /3 + y sin cos p = 6 sin £. (6) 

Subtract (5) from (6), x (cos a sin £ — sin a cos /3) = b sin /3 — a cos /3. 

By [8], x sin (/3 — a) — b sin p — a cos /3. 

_ b sin /3 — a cos /3 



sin (/3 — a) 



190. Solve for x and ?/ the system 



sin x + sin ?/ = a, (1) 

cos x + cos y — b. (2) 

Transform (1) by [20], 2 sin i(x + y) cos | (x - y) = a. (3) 

Transform (1) by [22], 2 cos } (x + y) cos $(x - y) =b. (4) 

(5) 
(6) 



Divide (3) by ( 
By [2], 


4), tan i(x + y) = - 



sin £ (x + y) _ a 

cos | (x + y) b 

sin $(x + y) a 


Sauare. 


Vl - sin 2 i(x + y) b 
sin 2 £ (x + y) _ as 



l-sinH(x + 2/) 6 2 
6 2 sin 2 £ (x + y) = a 2 - a 2 sin 2 } (x + y). 
(a 2 + & 2 )sinH(x + ?/) = a 2 . 

sinH(aj + y) = _^_. 

sinj(x + y)=: (7) 

Va 2 + 6 2 
Substitute in (3) the value of sin£(x -f ?/), 

cos|(x - ?/) = a. 
Va 2 + ft 2 



cosi(x -y) = i Va 2 + 6 2 . (8) 

From (5), a + y = 2 tan- 1 ?. (9) 



From (8), x - y = 2 cos- 1 J Va 2 + 6 2 . (10) 



teachers' edition. 221 



Add (9) and (10), and divide by 2, 



x = tan - 1 - + cos - 1 1 Va 2 + b 2 . 
b 



Subtract (10) from (9), and divide by 2, 



y — tan ~ 1: — cos - 1 \ Va 2 + b 2 . 
b 



191. Solve for r 


and the 


system 

r sin — a, 
r cos = b. 


Divide (1) by (2), 




tan = - • 
b 

.-. = tan- 1 -- 
b 


Square (1), 




r 2 sin 2 = a 2 . 


Square (2), 




r 2 cos 2 = 6 2 . 


Add (5) and (6), 


r 2 (sin 2 


+ cos 2 0) = a 2 + 6 2 . 
... r 2 = a 2 + B 2 . 
.-. r = Va 2 -h ft 2 . 



(1) 

(2) 
(3) 

(4) 

(5) 
(6) 



192. Solve for r and the system 

rshi(0 + a) = a, (1) 

r cos (0 + /3) = 6. (2) 

Expand (1) by [4], 

r sin cos a + r cos sin a = a. (3) 

Expand (2) by [6], . 

r cos cos £ — r sin sin /3 = 6. (4) 

Multiply (3) by cos 0, 

r sin cos a cos /5 -f r cos sin a cos p = a cos £. (5) 

Multiply (4) by sin or, 

- r sin sin a sin £ + r cos sin a cos = 6 sin a. (6) 
Subtract (6) from (5), 

r sin (cos a cos j8 -f sin a sin £) = a cos — 6 sin a. 

By [9], r sin cos (a — j8) = a cos £ — b sin a. 

. n a cos |3 — & sin a ._. 

.-. rsin0 = - • (7) 

cos (a — /3) 

Multiply (3) by sin /3, 

r sin cos a sin /3 + r cos sin a sin /3 = a sin j8. (8) 

Multiply (4) by cos a, 

- r sin cos a sin /3 + r cos cos a cos £ = & cos a. (9) 



222 PLANE TRIGONOMETRY. 

Add (8) and (9), 

r cos (sin a sin /3 + cos a cos (3) = a sin /3 -f b cos a. 
By [9], r cos cos (a — /3) = a sin £ + b cos a. 






a sin fl + 5 cos a /HM 

.-. rcos0 = r . (10) 

cos (a - |8) 

-r^. . i /*-v •. « a cos j8 — 6 sin a ,„„, 

Divide (7) by (10), tan = (11) 

a.sin j8 + 6 cos a 

_ acos/3 — 6 sin a ,_ 

.-. = tan- 1 (12) 

a sin j8 + b cos a 

From (1) and (2), r = = (13) 

w w sin(0 + a) cos(0 + /3) v ' 

Substitute in (13) the value of found in (12), 

a b 



. / . acos/3 — 6 sin a \ / . a cos 8 — b sin a \ 

sin ( tan— r h a ) cos ( tan - 1 - h p J 

\ a sin /3 -f 6 cos a / \ a sin j8 + 6 cos a / 



193. Solve for r, 0, and the system 





r cos sin = a, 


(1) 




r cos cos = 6, 


(2) 




r sin = c. 


(3) 


Divide (1) by (2), 


tan = - • 
b 


(4) 




.-. = tan- 1 -. 
b 


(5) 


Square (1), 


r 2 cos 2 0sin 2 = a 2 . 


(6) 


Square (2), 


r 2 cos 2 cos 2 = b 2 . 


(7) 


Add (6) and (7), 


r 2 cos 2 (sin 2 + cos 2 0) = a 2 + ft 2 . 






.-. r 2 cos 2 = a 2 -f 6 2 . 


(8) 




.-. r cos = Va 2 + 6 2 . 


(9) 


Divide (3) by (9), 


c 
tan — • 




Va 2 + b 2 






.-. = tan— 1 


(10) 




Va 2 + b 2 




Square (3), 


r 2 sin 2 = c 2 . 


(11) 


Add (11) and (8), 


r 2 (sin 2 + cos 2 0) = a 2 + b 2 + c 2 . 
.-. r 2 = a 2 + 6 2 + c 2 . 





r = Va 2 + b 2 + c 2 . 



teachers' edition. 



223 



194. Solve for x and y the system 

x sin 21° + y cos 44° = 179.70, (1) 

x cos 21° + y sin 44° = 232.30. (2) 

0.3584 x + 0.7193 2/ = 179.70. (3) 

0. 9336 x + 0. 6947 y = 232. 30. (4) 

3584 x + 7193 y = 1797000. (5) 

9336 x + 6947 y = 2323000. (6) 

4182528 x + 8394231 y = 2097099000. (7) 

4182528 x + 31122562/ = 1040704000. (8) 

5281975?/ = 1056395000. 

.-. y = 200. 
Substitute the value of y in (5), 

3584 x + 1438600 = 1797000. 

3584 x = 358400. 

.\ x = 100. 



Multiply (5) by 1167, 
Multiply (6) by 448, 
Subtract (8) from (7), 



195. Solve for x and y the system 

sin x — sin y = 0.7038, 
cos x — cos y — — 0.7245. 

Expand (1) by [21], 

2 cos i (x + y) sin £ (x - y) = 0. 7038. 
Expand (2) by [23], 

— 2 sin i (x + y) sin £ (x — y) = — 0. 7245. 

Divide (4) by (3), tan i (x + y) = 0,/245 . 
v ; J v " *v -r*/ 0.7038 

sin|(x + 2/) _ 0.7245 

cosi(x + 2/) ""0.7038* 

sin£(x + 2/) _ 0.7245 



(1) 
(2) 

(3) 

(4) 
(5) 



Vl -sin 2 £(x + 2/) ' 7038 
sin 2 £(x + 2/) _ 0.7245 2 
1 - sin 2 £ (« + 2/) ~ 0.7038 2 ' 
0.7038 2 sin 2 £(x + y) = 0.7245 2 - 0.7245 2 sin 2 £(x + y). 
(0 .7038 2 + 0.7245 2 ) sin 2 |(x + 2/) = 0.7245 2 . 
V0.7038 2 + 0.7245 2 sin i(x + y) = ±0.7245. 

. w - , ± 0.7245 

sin i (x + V) = ■ 



sin|(x + y)-. 



V0.7038 2 + 0.7245 2 
±0.7245 



1.0100667 
sini(x + y) = ±0.7172. 



(6) 



224 PLANE TRIGONOMETRY. 

Substitute in (4) the value of sin \(x + 2/)> 
2(± 0.7172) sin i(x - y) = 0.7245. 

. ', x 0.7245 

sm i(x - y) = - 






±1.4344 

sin i(x- y) = ±0.5051. (7) 

From (6), i(x + y) = 45° 50', 134° 10', 225° 50', or 314° 10'. (8) 

From (7), }(x-y) = 30° 20', 149° 40', 210° 20', or 329° 40'. (9) 

Add (8) and (9), x = 76° 10', 283° 50', 436° 10', or 643° 50'. 

Subtract (9) from (8), y = 15° 30', 344° 30', 15° 30', or 344° 30'. 

.-. x = 76° 10' or 283° 50', 
y = 15° 30' or 344° 30'. 
But the values x = 283° 50', y = 344° 30' do not satisfy the given 
system of equations. 

.-. x = 76° 10' ; y = 15° 30'. 



196. Solve for r and the system 

r sin = 92.344, (1) 

y cos (9 = 205.309. (2) 

qo Q44 
Divide (1) by (2), tan 6 = _^Ll^Z - 0.4498. 

w J K h 205.309 

.-. 6> = 24° 13' or 204° 13'. (3) 

92.344 92.344 

From (1), r = = = ± 225.12. 

v " sin 6 ± 0.4102 

.-, = 24° 13', r = 225.12; 

r = 204° 13', r = - 225.12. 

197. Solve for r and the system 

r sin (0 - 19° 18') = 59.4034, (1) 

r cos (0 - 30° 54') = 147.9347. (2) 

Expand (1) by [8], 

r sin cos 19° 18' - r cos sin 19° 18' = 59.4034. (3) 

Expand (2) by [9], 

r cos cos 30° 54' + r sin sin 30° 54' = 147.9347. (4) 

Multiply (3) by sin 30° 54', 

r sin sin 30° 54' cos 19° 18' - r cos sin 19° 18' sin 30° 54' 

= 59. 4034 sin 30° 54'. (5) 

Multiply (4) by cos 19° 18', 

r sin sin 30° 54' cos 19° 18' + r cos cos 19° 18' cos 30° 54' 

= 147.9347 cos 19° 18', (6) 



teachers' edition. 225 

Subtract (5) from (6), 

r cos 8 (cos 19° 18' cos 30° 54' + sin 19° 18' sin 30° 54') 

= 147.9347 cos 19° 18' - 59.4034 sin 30° 54'. 
By [9], r cos cos (30° 54' - 19° 18') 

= 147.9347 cos 19° 18' - 59.4034 sin 30° 54'. 

. 147.9347 cos 19° 18' - 59.4034 sin 30° 54' 

.-. rcos0 = (7) 

cos 10° 36' v ' 

Multiply (3) by cos 30° 54', 

r sin B cos 30° 54' cos 19° 18' - r cos 6 cos 30° 54' sin 19° 18' 

= 59.4034 cos 30° 54'. (8) 

Multiply (4) by sin 19° 18', 

r sin 6 sin 30° 54' sin 19° 18' + r cos d cos 30° 54' sin 19° 18' 

= 147.9347 sin 19° 18'. (9) 

Add (8) and (9), r sin 6 (cos 30° 54' cos 19° 18' + sin 30° 54' sin 19° 18') 
= 147.9347 sin 19° 18' + 59.4034 cos 30° 54'. 

By [9], r sin $ cos (30° 54' - 19° 18') 

= 147.9347 sin 19° 18' + 59.4034 cos 30° 54'. 

. m 147.9347 sin 19° 18' + 59.4034 cos 30° 54' „.. 

.-. r sin 6 = (10) 

cos 10° 36' v ' 

t^. . /-.m-u ,*x * n 147.9347 sin 19° 18' + 59.4034 cos 30° 54' 

Divide (10) by (7), tan0 = 

v ' ' 147.9347 cos 19° 18' - 59.4034 sin 30° 54' 

_ 147.9347 x 0.3305 + 59.4034 x 0.8581 

~ 147.9347 x 0.9438 - 59.4034 x 0.5135 

_ 48.89241835 + 50.97405754 

" 139.62076986 - 30.50364590 

99.86647589 



109.11712396 
= 0.9152. 
.-. 6 = 42° 28' or 222° 28'. (11) 

Substitute in (1) the value of 6 found in (11), 

r sin (42° 28' - 19° 18') = 59.4034 or r sin (222° 28' - 19° 18') = 59.4034. 
r sin 23° 10' = 59.4034 or r sin 203° 10' = 59.4034. 
59.4034 59.4034 



r = 



sin 23° 10' sin 203° 10' 

59.4034 59.4034 

or r — - 



0.3934 -0.3934 

r = 151 or r = — 151. 

.-. 6 = 42° 28', r = 151 ; 
or $ = 222° 28', r = - 151. 



226 PLANE TRIGONOMETRY. 

198. Solve for r, 0, and the system 

r cos cos = - 46. 7654, (1) 

rsin0cos0 = 81, (2) 

r sin = - 54. (3) 

81 
Divide (2) by (1), tan <p = _ ^ ^ = - 1.7320. 

.-. = 120° or 300°. 
.-. cos = T i- (4) 

Substitute in (1) the value of cos found in (4), 
r(=Fi)cos0 = - 46.7654. 

r cos d = ± 93.5308. (5) 

Divide (3) by (5), tan = — ~ 54 = T 0.5773. 

w J w ±93.5308 

.-. = 150° or 330° ; or 30° or 210°. 

.-. sin - i or - i ; \ or - -J. (6) 

Substitute in (3) the value of sin found in (6), 
r(±i) = -54. 
.-. r = =F 108. 
.-. r = 108, = 120°, = 330° ; r = 108, = 300°, = 210° ; 
r = - 108, = 120°, = 150° ; or r = - 108, = 300°, = 30°. 

199. Eliminate from the system 

x = r(0 -sin0), (1) 

y - r{\ — cos0). (2) 

Now 1 — cos = vers 0. (3) 

Substitute in (2) the value of 1 — cos found in (3), 

y = r vers 0. 

v 
vers = - • 
r 

.-. = vers- 1 -. (4) 

From (2), 1 -cos0 = ^. 

1_ Vl-sin20=X 
r 

1 - ? = Vl~sin2 0. 
r 

1 _^ + ^! = i_ sin2< ,. 

r r 2 



TEACHERS EDITION. 

r r 2 r 2 

.-. sin 6 = ± - V2 ry — y 2 . 



221 



(SJ 



Substitute in (1) the value of d found in (4), and the value of sin 

found in (5), 

x = r vers- * - T V2 ry - y 2 , 
r 



or x 



= ± V2 ry - y 2 + : 



Exercise XXV. Page 121. 

1. Given logi 2 = 0.30103, logi 3 = 0.47712, log 10 7 = 0.84510; find 
logio 6, logio 14, logio 21, logio 4, logio 12, logio 5, logio i, logioi, logio J, 



logio ft. 




logio 6 = logio 2 + logio 3 
logi 2 =0.30103 
logi 3 = 0.47712 
.-.logio 6 =0.77815 


logio 14 = logi 2 + logio7 
logio 2 =0.30103 
logio 7 =0.84510 
.-.logio 14 = 1.14613 


logio 21 = logio 3 + logio 7 
logio 3 = 0.47712 
logio 7 = 0.84510 
.-. logi 21 = 1.32222 


logio 4 = 2 logio 2 

logio 2 = 0.30103 

2 

.-. logio 4 = 0.60206 


logio 12 = log 10 3 + logi 4 
logio 3 = 0.47712 
logio 4 = 0.60206 
.-.logio 12 = 1.07918 


logio 5 = logio 10 -logio 2 
logio 10 = 1.00000 
logio 2 = 0.30103 
.-. logio 5 = 0.69897 



logio i 
logio 1 
logio 2 

logio i 

logio J 
logio 7 
logio 3 2 

■ logio | 



logio 2 



= logio 1 
= 0.00000 
= 0.30103 
= 1.69897 

= logio 7 - logio 1 
= 0.84510 
= 0.95424 

= 1.89086 



logio i 
logio i 



= 2 logio i 
= 1.69897 

2 

logioi =1.39794 



logio H 
logio 21 



= logi 21 - logi 20 
= 1.32222 



logio (10x2) = 1.30103 
logioM =0.02119 



2. With the data of Example 1 ; find 

logs 10, log 2 5, log 3 5, log 7 £, logssfg. 



228 PLANE TRIGONOMETRY. 



, og2 10 = lggiolO = 1 =88al9> 
logi 2 0.30103 

log 2 5 = ^ 5 = 0_^9897 = 
logi 2 0.30103 

log 3 5 =^l = °^ = 1.4660. 
5 logi 3 0.47712 

log 7 i =^ ^~ 0.30103 ^_ 

5 ™ log 10 7 0.84510 

logsrfs =21og 5 3-31og 5 7 

== 21og 1 o3-31og 10 7 

logi 5 

_ 0.95424 - 2.53530 

0.69897 

= - 2.2620. 

3. Given logi e = 0.43429 ; find 

log e 2, log e 3, log e 5, log e 7, log e 8, log e 9, log e f, log e |, log e ff, log e %V 

log e 2 = **£* = ^1^ = 0.69315. 
& logioe 0.43429 

log e 3 = !f^ = 9^712 
5 logioe 0.43429 

loge5 = 1 ^ = » 7 = 1.60945. 
8 log 10 e 0.43429 

loge7 = ^ = o_^I_o 

5 log w e 0.43429 

! = 3xlog 10 2 = : 90309 = 

s logioe 0.43429 

log e 9 = 2xlogio3 = 0_ : 954j4 =219724| 

6 logioe 0.43429 
logef = log e 2 - log e 3 = - 0.40547. 
log e | = 21og e 2-log e 5=-0.22315. 
log e |f = log e 5 + log e 7 - 3 log e 3 = 0.25952. 

10g e ^ = lQ ge 7 ~ ( lQ ge5 + log e 3 + 2 l0g e 2) 

= - 2. 14845. 

4. Find x from the equations 5* = 12, 16* = 10, 27* = 4. 

5* = 12. .\x logio5 = logi 12. 

aj = iogioi2 = Lom8 = 1 . 54396 . 

logi 5 0.69897 
16* = 10. .-. x logi 16 = logiolO. 



TEACHERS EDITION. 



229 



X = 



logiolO 1.00000 



= 0.83048. 



Iogiol6 1.20412 

27* = 4. .-. x logi 27 = logi 4. 

logi 4 0.60206 



x =■ 



logi 27 1.43136 



= 0.42062. 



Exercise XXVI. Page 126. 

1. Calculate to five places of decimals log e 3. 

In the case of log e 3 the calculation is carried out below to ten places, 
for use in Example 4. 
In the formula 

z+1 ./ 1 1 1 



10g e 



-(i 



+ 



+ 



let 

Then 
and 



2z + l 3(2z + l) 3 5(22 + l> 



r. + ' 



g + 1 

z 

lOge 3 

2 
4 
4 
4 
4 
4 
4 
4 
4 
4 
4 
4 
4 
4 
4 
4 
4 



= 3, and 2 2 + 1 = 2 ; 



.(1 

\2 



+ 



1 



3 x 2 3 
2.00000000000 



X 25 



+ 



1.00000000000 - 1 = 1.00000000000 
0.25000000000 -*- 3 = 0.08333333333 
0.06250000000 - 5 = 0.01250000000 
0.01562500000 ~ 7 = 0.00223214286 
0.00390625000 -*- 9 = 0.00043402778 
0.00097656250 ~ 11 = 0.00008877841 
0.00024414062 - 13 = 0.00001878005 
0.00006103515 -*- 15 = 0.00000406901 
0.00001525879 - 17 = 0.00000089758 
0.00000381470 -f- 19 = 0.00000020077 
0.00000095367 - 21 = 0.00000004541 



0.00000023842 - 23 = 0.00000001037 
0.00000005960 -f- 25 = 0.00000000238 
0.00000001490 ~ 27 = 0.00000000055 



0.00000000372 -*- 29 = 0.00000000013 



0.00000000093 - 31 = 0.00000000003 



0.00000000023 - 33 = 0.00000000001 
1.09861228867 
/.loge 3 = 1.0986122886. 



230 



PLANE TRIGONOMETRY. 



2. Calculate to five places of decimals log e 5. 

For use in this example and in succeeding examples let us first calcu- 
late to eight places of decimals the value of log e 2. 

Let z = 1. 

Then z + 1 = 2, and 2z + 1 = 3 ; 

1 1 



and 



Let 
Then 



and 



log e 2 = 2(i 



3 + 3 x 3» 



+ 



5 x 3 5 



+ ■ 



2.000000000 



0.666666667 + 1 = 0.666666667 
0.074074074 h- 3 = 0.024691358 

0.001646091 
0.000914495 -=- 7 = 0.000130642 
0.000101611 ~ 9 = 0.000011290 

0.000001026 

0.000000096 
= 0.000000009 

0.000000001 



0.000011290 



0.000001254 



0.1)00000015 



1 

3: 

5 : 

7: 

9: 

11 

13: 

15: 

17 



0.693147180 
.♦.log e 2 = 0.69314718. 

z = 4. 

2 + 1 = 5, and 2 z + 1 = 9 ; 
5 ./l . 1 



loge 



■<i 



9 + 3 x 93 



+ 



5 x 9 5 



+ 



2.000000 



0.222222 -f- 1 = 0.222222 



0.024691 



0.002743 - 3 = 0.000914 



0.000305 



0.000034 + 5 = 0.000007 
0.223143 



log e J = 0.22314. 

loge 5 = 0.22314 + loge 4 

= 0.22314 + 2 x loge 2 
= 0.22314 + 2 x 0.69315 
= 1.60944. 



TEACHERS 5 EDITION. 



231 



3. Calculate to five places of decimals log e 7. 



Let 
Then 



and 



2 + 1 = 7, and2z + 1 = 13; 

1 1 



& 6 \ll 



+ 



13 3 x 133 5 x 135 



13 
13 
13 
13 
13 



2.000000 



0.153846 -f- 1 = 0.153846 



0.011834 



0.000910 -f- 3 = 0.000303 



0.000070 



0.000005 -4- 5 = 0.000001 
0.154150 



loge f=0.15415. 

loge 7 = 0.15415 + log e 6 

= 0.15415 + log e 2 + loge 3 

= 0.15415 + 0.693147 + 1.098612 

= 1.94591. 



4. Calculate to ten places of decimals log e 10 

Let z = 9. 

Then z -f 1 = 10, and 2 z + 1 = 19 ; 
10 = 2fJU 1 



and 



log e 



9 



+ ■ 



\19 3 x 19 3 5 x 19 5 



7 + 



19 


2.00000000000 


■1 

-3 
-5 

-7 




19 


0.10526315789 -f 


= 0.10526315789 


19 


0.00554016620 




19 


0.00029158769 - 


= 0.00009719590 


19 


0.00001534672 




19 


0.00000080772 -? 


= 0.00000016154 


19 


0.00000004251 






0.00000000224 -f 


= 0.00000000032 
0.10536051565 



loge- 1 / = 0.1053605156. 

log e 10 = 0.1053605156 + 2 log e 3 

= 2.3025850930. 



232 PLANE TRIGONOMETRY. 

5. Calculate to five places of decimals logio 2, logio e, logio 11. 

log 10 2 =*«•■= ^?^ = 0.30103. 
log e 10 2.302585 

log 10 e = J^i = 1 = 0.43429. 

log e 10 2.302585 

To calculate logio 11, let z = 10. 

Then z -f 1 = 11, and 2 \z + 1 = 21 j 

and logio H = 2 logio e ( 1 1 h • • • J • 

10 5 \21 3 x 213 5 x 21 5 ) 

2.000000 



21 
21 
21 



0.095238 -f- 1 = 0.095238 



0.004535 



0.000216 -f- 3 = 0.000072 
0.095310 
•"■ lo Sio H = 0.09531 x logio e 
= 0.09531 x 0.43429 
= 0.04139. 
logio 11 = 0.04139 + logio 10 
= 1.04139. 



Exercise XXVII. Page 128. 

1. Given it — 3.141592653589; compute sin l 7 , cos 1', and tan V to 
eleven places of decimals. 
. The circular measure of V is 

-*- = 3 - 141592653589 = 0.0002908882 +, 
10800 10800 

the next figure being or 1. 

Again, taking the value of sin V as computed in the text-book 7 
0.00029088 +, we have 



cos V > VI - (0.00029089) 2 
>Vl- 0.000000084617 
>Vo. 999999915383 
> 0.999999957691. 

Also cos V < Vl - (0.00029088) 2 

< Vl - 0.000000084611 
<Vo. 999999915389 

< 0.999999957694. 



233 

Hence, cos V = 0.99999995769, correct to eleven decimal places. 

By Sect. XLV, sin x > x cos x. 

.-. sin 1 7 > 0.0002908882 x 0.99999995769 

> 0.0002908882 (1 - 0.00000004231) 

> 0.0002908882 - 0.000000000012 

> 0.00029088818. 

Therefore, sin V lies between 0.00029088818 and 0.00029088821. That 
is, correct to nine places of decimals, 

sin V = 0.000290888, 
the next two figures being 18, 19, 20, or 21. 

Repeating the process, beginning with the last value of sin 1', the com- 
putation can be carried still further. To eleven places, 
sin V = 0.00029088820. 

From the values of sin V and cos V we have 

* t/ sinr 
tan V = . 



cos V 

_ 0.00029088820 
"0.99999995769 
= 0.000290888212. 

2. Given # = 3.141592653589; compute sin 2' by the same method, 
and also by the formula sin 2 x = 2 sin x cos x. Carry the operations to 
nine places of decimals. Do the two results agree ? 
The circular measure of 2' is 

^ = 3.141592653589 = 0Q00581776 
5400 5400 

Hence, sin2 7 lies between and 0.0005817765, 



and cos 2' > Vl - (0.0005817765) 2 

> Vl - 0.0000003384638959 



>V0. 9999996615361040 

> 0.999999830768. 
But sin x > x cos x. 

.-. sin 2'> 0.0005817764 x 0.99999983076 

> 0.0005817764(1 -0.00000016924) 

> 0.0005817764 - 0.000000000098 

> 0.0005817763. 

Hence, sin 2' = 0.000581776, correct to nine decimal places. 



234 



PLANE TRIGONOMETRY. 



Again, sin 2' = 2 sin V cos V 

= 2 x 0.00029088820 x 0.99999995769 
= 0.00058177640 (1 - 0.00000004231) 
= 0.00058177640 - 0.000000000025 
= 0.000581776+. 

The two methods, therefore, agree to at least nine decimal places. 



3. Given it = 3.141592653589 ; 
compute sin 1° to four places of 
decimals. 

The circular measure of 1° is 
it _ 3.14159265 3589 
180~ 180~ 

Hence, 



= 0.01745329. 



cos 1° > Vl - (0.01746) 2 
>V0. 9996951 5 

> 0.999847. 
sin x > x cos x. 

/. sin 1°> 0.017453 x 0.999847 ' 

> 0.017453 (1-0.000153) 

> 0.017453- 0.0000027 

> 0.0174503. 

Hence, to four decimal places, 
sin 1° = 0.0175. 

4. From the formula 

x 
cos x = 1 — 2 sin 2 - 



show that cos x > 1 

2 

By Sect. XLV, sin x < x. 

. x x 
.*. sin-<- 
2 2 

sm 2 - < — • 

2 4 

l-2sin 2 ?>l--. 

2 2 



.-. COS X > 1 



X' 



5. Show by aid of a table of 
natural sines that sin x and x agree 



to four places of decimals for all 
angles less than 4° 40'. 

The circular measure of 4° 40', or 
280', is 

280 n _ 7 it 
10800 ~ 270 

_ 7 x 3.J41592653589 
~ 270 

= 0.0814487. 
The circular measure of 4° 4r is 
0.0814487 + 0.0002909 = 0.0817396. 
From a table, 

sin 4° 40' = 0.0814. 
sm4°4r = 0.0816. 
Hence, sin x and the circular meas- 
ure of x agree for 4° 40 r , and there- 
fore for all smaller angles to four 
decimal places ; but they differ for 
larger angles. 

6. If the values of log x and log 
sin x agree to five decimal places, 
find from a table the greatest value 
x can have. 

Let x be expressed in seconds. 
Then its circular measure is 



648000 
and its logarithm is 

log x" + (log it - log 648000) 
= logx" + (0.49715 - 5.81158) 
= logx // - 5.31443 
= log x" + 4.68557 -10. 



teachers' edition. 235 

But from the explanation preceding Table IV, if we remember that 

log sines are given in the table increased by 10, we have 

log sin x + 10 = log x" -f S. 

.-. log sin x = log x" + S — 10. 

Hence, if, for five decimal places, log sin x = log x, we have 

log x" + 4.68557 - 10 = log x" + S - 10. 

.-. S = 4.68557. 

But, the greatest angle for which this value of S can be used is given 

in the table as 2409 // . 

Hence, the greatest angle for which log x and log sin x agree to five 

decimal places is 

F 2409" = 40' 9". 

Exercise XXVIII. Page 130. 

1. Compute the sine and cosine of 6 / to seven decimal places. 
From Prob. 2, Ex. XXVII, 

sin 2' = 0.000581776. 
Also, from Prob. 1, Ex. XXVII, 

cos r = 0.999999958, 
and sin V = 0.000290888. 
Hence sin 3' = 2 sin 2 / cos V — sin V 

= 2 x 0.000581776 (1 - 0.0000001) - 0.000290888 

= 2x 0.000581776 - 0.000290888 

= 0.000872664. 
cos 2' - 2 cos 2 V - 1 

= 2(0.999999958) 2 -l 

= 2 x 0.999999916-1 

= 0.999999832. 
cos 3' = 2 cos 2' cos V — cos V 

= cos V (2 cos 2' — 1) 

= 0.999999958(2 x 0.999999832 - 1) 

= (1 - 0.000000042) (0.999999664) 

= 0.999999622. 
By [12], sin 6' = 2 sin 3 7 cos 3' 

= 2 x 0.000872664 x 0.999999622 

= 0.001745327. 
By [13], cos 6' = 2 cos 2 3 7 - 1 

= 2(0.999999622) 2 - 1 

= 2 x 0.999999244-1 

= 0.999998488. 



236 PLANE TKIGONOMETRY. 

2. In Formula (1) let y = 1°. Assuming sin 1° = 0.017454 + , cos 1° ! 
0. 999848 + , compute the sine and cosine of two degrees. 

sin 2° = 2 sin 1° cos 1° 

= 2 x 0.017454 x 0.999848 

= 0.034902. 
cos 2° = 2cos 2 l°- 1 

= 2 x (0. 999848) 2 - 1 

= 2 x (0.999696) -1 

= 0.999392. 



3. In Formula (1) let y = 1°. Assuming sin 1° = 0.017454 + , cos 1° i 
0. 999848 + , compute the sine and cosine of three degrees. 

sin 3° = 2 sin 2° cos 1° - sin 1° 

= 2 x 0.034902 x 0.999848 - 0.017454 

= 0.052339. 
cos 3° = (2 cos 2° - 1) cos 1° 

= 0.998784 x 0.999848 

= 0.998632., 



4. In Formula (1) let y = 1°. Assuming sin 1° = 0. 017454 + , cos 1° = 
0. 999848 + , compute the sine and cosine of four degrees. 

sin 4° = 2 sin 3° cos 1° - sin 2° 

= 2 x 0.052339 x 0.999848 - 0.034902 

= 0.069760. 
cos 4° = 2 cos 3° cos 1° - cos 2° 

= 2x 0.998630 x 0.999848 - 0.999392 

= 1.996956-0.999392 

= 0.997564. 



5. In Formula (1) let y = 1°. Assuming sin 1° = 0.017454 + , cos 1° = 
0. 999848 + , compute the sine and cosine of five degrees. 

sin 5° = 2 sin 4° cos 1° - sin 3° 

= 2 x 0.069760 x 0.999848 - 0.052339 
= 0.087160. 

cos 5° = 2 cos 4° cos 1° - cos 3° 



= 2 x 0.997564 x 0.999848 - 0.998632 
= 1.994825-0.998632 
= 0.996193. 



teachers' edition. 



237 



Exercise XXIX. Page 135. 



1. Find the six 6th roots of — 1 ; 
of + 1. 

- 1 = cos 180° + i sin 180°. 
+ 1 = cos 0° + 1 sin 0°. 

Hence, the six 6th roots of — 1 
are 

cos 30° + i sin 30° = 8 + z , 



cos 90° + i sin 90° = i, 
cos 150° + i sin 150° = 



-V3 + i 



cos 210° + i sin 210° = 
cos 270° + z sin 270° = 
cos 330° + i sin 330° = 



-V5- 

2 

V3-i 



The six 6th roots of + 1 are 
cos 0° + i sin 0° 
cos 60° + i sin 60° = ■ 

cos 120° + i sin 120° = 



cos 180° + i sin 180° = - 1, 

cos 240° + i sin 240° = nLz H? , 



cos 300° + i sin 300° = 



1 -V-3 



2. Fiod the three cube roots of i. 

i - cos 90° + i sin 90°. 
Hence, the three cube roots of i 



cos 30° + i sin 30° = 



cos 150° + i sin 150° = 



V3 + i 

* 

2 

-V3 + i 



cos 270° + i sin 270° = — i 

3. Find the four 4th roots of — i. 
-i = cos 270° + i sin 270°. 





Hence, the four 4th roots of — i 


+ 1, 


are 


l+V-3 


cos 67i° + £ sin 67£°, 


2 


cos 157£° + * sin 157i°, 


_l + V-3 


cos 247i° + i sin 247i°, 


2 


cos 337i° + t sin 337i°. 



4. Express sin 4 6 and cos 4 in terms of sin 6 and cos 0. 

By Sect. XLVII, 

4x3x2 
sin 4 = 4 cos 3 sin 0* , cos sin 3 



and 



= 4 cos 3 sin — 4 cos sin 3 ; 

4v^ 4x3x2x1 
cos40 = cos 4 r— cos 2 0sin 2 + r— — sin 4 



li 



= cos 4 0-6 cos 2 sin 2 + sin 4 0. 



238 PLANE TRIGONOMETRY. 



Exercise XXX. Page 137. 
1. Verify by the series just obtained that sin 2 x -f cos 2 x = 1. 

X 3 X 5 X 7 

sin x = x 1 h • • • 

6 120 5040 

X 2 X 4 X 6 X 8 

cos x = 1 1 h 



2 24 720 40320 

. x 4 2x 6 x 8 

.-. sin 2 x = x 2 h • • • 

3 45 315 

x 4 2x 6 x 8 
and ** B =I-* + -- — + — - 

.*. sin 2 x + cos 2 x = 1. 



2. Verify by the series just obtained that sin (— x)-= — sin x and 

cos ( — x) = cos x. 

The series for the sine consists entirely of odd powers of x and, there- 
fore, changes its sign with x ; while the series for the cosine consists 
entirely of even powers, and is unchanged when x changes its sign. 

3. Verify by the series just obtained that sin 2 x = 2 sin x cos x. 

. _ a (2x) 3 (2x) 5 (2x) 7 
sin 2 x = 2 x - — -?- + i — - - ^— J- + • • • 



Also 







6 


120 5040 




= 2x 


4x 3 

~~3~ + 


4x 5 8x 7 
T5" ~315 + " 




V 3 


2x 5 4x 7 
+ T5~~315 + 


sin x cos x 


= x — 


2x 3 2 

~3~ + ~ 


x 5 4x 7 
5~~315 + '" 


.*. sin2x 


= 2 sin x cos x 





By Prob. 1, 


• o x 4 2x 6 x 8 , 

sm 2 x = x 2 h • h • • • 

3 45 315 


.-. 1- 


c • , « 2x 4 4x 6 2x 8 

- 2 sm 2 x = 1-2 x 2 H h 

3 45 315 




(2x) 2 (2x) 4 (2x)6 (2x) 8 
2 24 720 40320 




= cos 2 x. 



TEACHERS' EDITION. 



239 



5. Find the series for sec x as far as the term containing the 6th power 
of x. 

/ x 2 x 4 x 6 \ 

V ~2~ + 24~720 + ") 



secx = — — = 1 -=- ( 1 

cosx 2 24 720 



« x 2 5x 4 61 x 6 

= 1 4- h 1 h •• 

2 24 720 



6. Find the series for x cot x, noting that x cot x = cos x. 

sinx 

x 3 x 5 x 7 

x cos x = x 1 h • • • 

2 24 720 

x 3 x 5 x 7 

sin x = x 1 h • • • 

6 120 5040 

X cos X X 2 X 4 2 X 6 

sinx 3 45 945 ~ 

7. Calculate sin 10° and cos 10° to five places of decimals. 



The circular measure of 10° is — 

18 



Hence 



sin 10 ( 



tc 1 / it ' 
18 ~~6\18> 



18/ 120 V 18/ 

, An . 1 / TC \ 2 1 / 7T\ 4 1 / it \ 6 

cos 10° = 1 - - ( — ) + — ( — ) ( — ) + 

2V18/ 24X18/ 720 \ 18/ 



Taking it = 3.141592653589, we find 
tc = 3.141593, 

^ 2 = 9.869604, 

?r 3 = 31.006277, 

** = 97.409091, 

7t* = 306.019685, 



18 



7T 2 



2 x 18 2 


7T 3 


6 x 18 3 


7t* 


24 x 18 4 


7T 5 


120 x 18 5 



0.174533, 
= 0.015231, 
= 0.000886, 
= 0.000039, 
= 0.000001. 



.-. sin 10° = 0.174533 - 0.000886 + 0.000001 
= 0.173648. 
cos 10° = 1 - 0.015231 + 0.000039 

= 0.984808. 

Note. The powers of it need be computed only once, and can then be 
used for finding the functions of all angles. 



240 PLANE TRIGONOMETRY. 

8. Calculate tan 15° to five places of decimals. 

Tt 

The circular measure of 15° is — 

12 

tt i co Tt 1 / Tt\* 2 / Tt\*> 17 / Tt \ 7 

Hence tan 15° = h - ( — ) H ( — ) H ( — ) -\ 

12 3X12/ 15X12/ 315X12/ 

it = 3.141593, — = 0.261799, 

?r 3 = 31.006277, — **— - = 0.005981, 

3 x 123 

n* = 306.019685, 2 ^ = 0.000164, 

15 x 125 

7T? = 3020.293227, — *' ** = 0.000005. 

315 x 127 

.-, tan 15° = 0.267949. 

9. From the exponential value of cos x show that 

cos 3x = 4 cos 3 x — 3 cos x. 

By Sect. XL VIII, cosx = i(e xi + e-»). 

.-. cos 3 x = i (e Sxi + e~ Sxi ) 

— %(e xi -f e~ xi ) (e 2xi - 1 + e- 2a *) 
= cosx [{4 x i(e xi + e-^')} 2 - 3] 
= cosx (4 cos 2 x — 3) 
'= 4 cos 3 x — 3 cosx. 

10. From the exponential value of sinx show that 

sin 3 x = 3 sin x — 4 sin 3 x. 

By Sect. XL VIII, sin 3 x = — (e 3 ** - e~ 3 ^') 

2i 

= __ (exi _ e -xi) (&xi + l -j- e-2^) 



= sin x ["(- 4) | — (e™ - e-*»") } + 3 ] 



= sin x (— 4 sin 2 x + 3) 
= 3 sin x — 4 sin 3 x. 



SPHERICAL TRIGONOMETRY. 






Exercise XXXI. Page 142. 



1. The angles of a triangle are 
70°, 80°, and 100°. Find the sides 
of the polar triangle. 

Given A = 70°, B = 80°, C = 100° ; 
to find a\ b', of. 



a' = 180° - 
b' = 180° ■ 
& = 180° - 



70° = 110°. 

80° = 100°. 

100°= 80°. 



2. The sides of a triangle are 40°, 
90°, and 125°. Find the angles of 
the polar triangle. 

Given a = 40°, b = 90°, c = 125° ; 
required A\ B', C. 

^' = 180°- 40° = 140°. 
B' = 180° - 90° = 90°. 
C =* 180° - 125° = 55°. 



3. Show that, if a triangle has 
three right angles, the sides of the 
triangle are quadrants. 

Every vertex is the pole of the 
opposite side. Every side is, there- 
fore, 90°. 



4. Show that, if a triangle has 
two right angles, the sides opposite 
these angles are quadrants, and the 
third angle is measured by the num- 
ber of degrees in the opposite side. 

Let ABC be the triangle, and 

B = C = 90°. 

Then A is the pole of a. There- 
fore, b and c are quadrants, and 
the angle A is equal to the side BC 
measured in degrees. 

5. How can the sides of a spheri- 
cal triangle, measured in degrees, 
be found in units of length, when 
the length of the radius of the sphere 
is known? 

Since the sides of the triangle are 
arcs of great circles, every degree of 
arc is T ^ of the circumference of a 

great circle, or , where r is the 

' 360 
radius of the sphere. Hence, to find 
the length of a side, multiply its 

. , , 2 7tr 7tr 

measure in degrees by or 

5 J 360 180 



6. Find the lengths of the sides of the triangle in Example 2 if the 
radius of the sphere is 4 feet. 

241 



242 



SPHERICAL TRIGONOMETRY. 



a = 40° = 40 x ^-^feet = —feet. 
180 9 

b = 90° = 90 x ?-2Li feet = 2 tt feet. 
180 

c = 125° = 125 x ^^ifeet = —feet. 
180 9 






Exercise XXXII. Page 146. 



1. Show, by aid of Formula [38] , 
p. 144, that the hypotenuse of a 
right spherical triangle is less than 
or greater than 90°, according as the 
two legs are alike or unlike in kind. 

By [38], cos c = cos a cos b. 

If a and b are both < 90° or both 
> 90°, cos a and cos b have the same 
sign. Hence, cose is positive, and 
c<90°. 

But if a and b are unlike in kind, 
cos a and cos b have opposite signs. 
Hence, cos c is negative, and c > 90°. 

2. Show, by aid of Formula [41], 
that in a right spherical triangle 
each leg and the opposite angle are 
always alike in kind. 

By [41], cos .4 = cos a sin B. 

Now JB<180°. 

.-. sin B is positive. 

Hence, the sign of cos A is same as 
the sign of cos a, and both must be 
greater than or both less than 90°; 
that is, alike in kind. 

3. What inferences may be drawn 
from Formulas [38]-[43] respecting 
the values of the other parts : 



(i) if c = 90° ; (ii) if a = 90 ; (iii) if 
c = 90° and a = 90°; (iv) if a = 90° 
and b = 90° ? 

(i) If c == 90°, 

[38] becomes = cos a cos b: 
.-. cos a or cos 6 = 0. 

.-. a or b = 90°. 
If a = 90°, 

[41] becomes 

cos A = x sin B = 0. 
.-. A *= 90°. 

Hence, from Prob. 4, Ex. XXXI, 
B = b. 



(ii) If 


a = 90°, 


[41] becomes 




cos A = x ski B. 




.-. A = 90°, 




c = 90°, 


and 


B = b. 


(iii) If 


c = 90°, 


and 


a = 90°, 


from (i) and 


(ii), A = 90°, 


and 


B = b. 


(iv) If 


a - 90°, 


and 


b = 90°, 


from (ii), 


c = 90°, 


and 


B = b = 90°. 



teachers' edition. 243 

4. Deduce from Formulas [38]-[43] and Formulas [18]-[23] the 
formula tan 2 \ b = tan \ (c — a) tan -£- (c + a). 

From [38], cos6 = C( 



By [18], tan 2 i& = 



By [23] and [22], 



cos a 
1 — cos b 
1 + cos b 
cos c 
cos a 
~ cos c 
cos a 
_ cos a — cos c 
~~ cos a + cos c 

_ — 2 sin i (a + c) sin £ (a — c) 
2 cos i (a + c) cos |(a — c) 
= — tan i (a + c) tan J (a — c) 
= tan i(c + a) tan £ (c — a). 



5. Deduce from Formulas [38]-[43] and Formulas [18]-[23] the 
formula tan 2 (45° — J A) = tan i (c — a) cot -} (c + a). 



From [39], 


sin A 


sin a 
sin c 


Now tan 2 (45° 


-iA) 


= tan 2 i(90°-^L) 
= cot 2 i(90° + ^) 


By [19], 




1 + cos (90° + A) 
~ 1 -cos(90° + ^i) 
1 — sin A 
1 + sin A 

sin a 

sin c 




sin a 
sin c 






sin c — sin a 



By [21] and [20], 



sin c + sin a 

2 cos | (c + a) sin -J- (c — a) 
2 sin £ (c + a) cos i(c — a) 
: cot £ (c + a) tan -J (c — a). 



244 SPHERICAL TRIGONOMETRY. 

6. Deduce from Formulas [38]-[43] and Formulas [18]-[23] the 
formula tan 2 £_B = sin (c — a) esc (c + a). 

-^ riAi ti tana 

From [40], cos B- 



By [18], tanH# = 



tan c 

1 — cos B 

1 -f- cos B 

tana 



tan c 



tan a 

tan c 
tan c — tan a 
tan c -f tan a 
sin c sin a 



cose cos a 



sin c sin a 



By [8] and [4], 



cos c cos a 
sin c cos a — cos c sin a 
sin c cos a + cos c sin a 
sin (c — a) 
sin (c + a) 
: sin (c — a) esc (c 4- a). 



7. Deduce from Formulas [38]-[43] and Formulas [18]-[23] the 
formula tan 2 i c = — cos (A + B) sec (A — B). 

^ r*^ COt A 

From [43], co 



By [18], tanHc: 



tanjB 
1 — cos c 
1 -|- cos c 
cot A 



tan B 



cot A 

tan B 
tan 5 — cot A 
tan 5 -f- cot A 
sin 5 cos ^4 
cos B sin u4 
sin B cos ^4 
cos 5 sin A 
sin .4 sin B — cos ^1 cos B 
sin ^4 sin J5 + cos A cos J5 



By [5] and [9], 



teachers' edition. 245 

_ - cos (A + B) 

~ cos (A - B) 

= - cos (A + B) sec (A - B). 



8. Deduce from Formulas [38]-[43] and Formulas [18]-[23] the 
formula tanH a = tan [I {A + B) - 45°] tan [£ (A - B) + 45°]. 

cos A 



From [41], co* 

By [18], tan 2 ia = 



By [20] and [21], 



sin B 
1 — cos a 

1 -f cos a 
cos A 

sin B 

cos A 

1 + 

sin B 

sin B — cos A 

sin B + cos ^1 

sin B + sin (J. - 90°) 

sin 5 - sin (J. - 90°) 

2sini(^4 + £-90°)cos£(£-J. + 90°) 



2cosi(^4 + B-90°)sm1z(B-A + 90°) 
= tan [i (A + B) - 45°] cot [i (5 - ^4) + 45°]. 
Now i(A - B) + 45° = 90° - ki(£ - ^L) + 45°]. 
.-. cot [i(B -A) + 45°] = tan [\(A - B) + 45°]. 

.-. tanHa = tan [}{A + B) - 45°] tan [\(A - B) + 45°]. 

9. Deduce from Formulas [38]-[43] and Formulas [18]-[23] the 
formula tan 2 (45° — -J- c) = tan |(^i — a) cot %(A + a). 






From [39], sin c = . 

sin A 

sin a 



.-. cos (90° - c) = 
By [18], tan 2 (45°-£c) = 



sin A 

1 -cos(90°-c) 
1 + cos(90°-c) 
sin a 



1 



sin A 



sin a 



sin^i 

sin ^ — sin a 

sin ^1 + sin a 



246 SPHERICAL TRIGONOMETRY. 

_ 2 cos i ( A + a) sin \{A — a) 






By [21] and [20], 



2 sin £ (^4 + a) cos -J- (A — a) 
= cot i (A + a) tan i (^1 — a). 



10. Deduce from Formulas [38]-[43] and Formulas [18]-[23] the 
formula tan 2 (45° — i b) = sin (A — a) esc (A + a). 

tan a 



From [42], sin 6 = 

.-. cos (90° -b) = 

By [18], tan 2 (45°-i&) = 



tan A 
tana 
tan J. 

1 - cos (90° - b) 
1 + cos (90° + b) 
tan a 



By [8] and [4], 



tan a 
tan J. 
tan A — tan a 
tan A -f tan a 
sin A sin a 
cos ^4. cos a 
sin i sin a 
cos A cos a 
sin A cos a — cos A sin a 
sin A cos a -f cos A sin a 
sin (J. — a) 
sin (^4 + a) 
: sin (^4 — a) esc (^4 + a). 



11. Deduce from Formulas [38]-[43] and Formulas [18]-[23] the 
formula tan 2 (45° - % B) = tan -J- (A - a) tan i(A + a). 

cos A 



From [41], sin B = 

.-. cos(90°-jB) = 

By [18], tan 2 (45° --£- B) = 



cos a 
cos A 
cos a 
1 - cos (90° - B) 



1 + cos (90°-£) 
cos ^4 



cos J. 






By [23] and [22], 



teachers' edition. 247 

_ cos a — cos A 
cos a + cos A 
_ — 2 sin i (a + ^4) sin -£ (a — ^1) 
2 cos J (a + A) cos £ (a — A) 
= — tan \(a + A) tan £ (a — A) 
= tan -J- (^L 4- a) tan i (-4. — a). 



Exercise XXXIII. Page 148. 

1. Show that Napier's Rules lead to the equations contained in 
Formulas [39], [40], [41], and [42]. 

sin a = cos (Co. c) cos (Co. A). 
.-. sin a = sin csin-4. [39] 

sin b = cos (Co. c) cos (Co. B). 
.-. sin b = sin c sin 5. [39] 

sin (Co. A) = tan 6 tan (Co. c). 

.-. cos A = tan 6 cot c. [40] 

sin (Co. 5) = tan a tan (Co. c). 

.-. cos 5 = tan a cot c. [40] 

sin (Co. A) = cos a cos (Co. B). 

.-. cos .A = cos a sin 5. [41] 

sin (Co. B) = cos 6 cos (Co. A). 

.-. cos B = cos bsinA. [41] 

sin 6 = tan a tan (Co. ^4). 
.-. sin b = tan a cot ^4. [42] 

sin a = tan 6 tan (Co. B). 
.-. sin a = tan b cot 5. [42] 

2. What will Napier's Rules become if we take as the five parts of the 
triangle the hypotenuse, the two oblique angles, and the complements of 
the two legs ? 

Each part will be replaced by its complement, and every function will 
be replaced by its complementary function. 
Therefore, Napier's Rules become 

Rule I. The cosine of any middle part is equal to the product of the 
cotangents of the adjacent parts. 

Rule II. The cosine of any middle part is equal to the product of the 
sines of the opposite parts. 



248 



SPHERICAL TRIGONOMETRY. 



Exercise XXXIV. Page 153. 



1. Solve the right triangle, given 
a = 36° 27', b = 43° 32' 31". 


By [38], cos c = 
log cos a = 


cos a cos b. 
9.90546 


log cos b = 


9.86026 


log cos c = 


9.76572 


.'. c = 


54° 20'. 


By [42], sin b = 
.*. tan a = 


tan a cot A. 
sin 5 tan A. 


.-. tan A = 


tan a esc 6. 


log tan a = 


9.86842 


log esc b = 


0.16185 


log tan A — 


10.03027 


A = 


46° 59' 43". 


By [42], sin a = 
.-. tan b = 


tan b cot B. 
sin a tan .B. 


.-. tan B = 


tan b esc a. 


log tan b = 


9.97789 


log esc a = 


0.22613 


log tan B = 


10.20402 


..B = 


57° 59 7 19". 



2. Solve the right triangle, given 
a = 86° 40', b = 32° 40 r . 

By [38], cos c = cos a cos 6. 
By [42], tan A = tan a esc b. 
By [42], tan 5 = tan b esc a. 

log cos a — 8.76451 
log cos b = 9.92522 
log cos c = 8.68973 

c = 87° IV 40". 

log tan a = 11.23475 
log esc b = 0.26781 
logtan^l = 11.50256 

A = 88° ir 58". 



log tan b = 9.80697 
log esc a = 0.00074 
log tan 5 = 9.80771 

B = 32° 42 7 39". 

3. Solve the right triangle, given 
a = 50°, b = 36° 54 7 49". 

By [38], cos c = cos a cos b. 

By [42], tan A = tan a esc 6. 

By [42], tan B = tan b esc a. 
log cos a = 9.80807 
log cos b - 9.90284 . 
log cos c = 9.71091 

c = 59° 4 7 26". 
log tan a = 10.07619 
log esc b = 0.22141 
logtan^. = 10.29760 

A = 63° W 13". 
log tan 6 = 9.87575 
log esc a = 0.11575 
log tan 5 = 9.99150 

B = 44° 26 7 22". 

4. Solve the right triangle, given 
a = 120° 10 r , b = 150° 59 7 44". 

By [38], cos c = cos a cos &. 

By [42], tan A = tan a esc 6. 

By [42], tan B = tan 6 esc a. 
log cos a = 9.70115 (n) 
log cos b = 9.94180 (n) 
log cos c = 9.64295 

c = 63° 55 r 43". 
log tan a = 10.23565 (n) 
log esc b = 0.31437 
log tan A = 10.55002 (n) 
A = 105° 44' 21". 
log tan b = 9.74383 (n) 
log esc a = 0.06320 
log tan B = 9.80703 (n) 
B = 147° W 47". 



TEACHERS EDITION. 



249 



5. Solve the right triangle, given 
c = 55° 9' 32", a = 22° 15' 7". 

By [38], cos b = cos c sec a. 

By [39], sin A = sin a esc c. 

By [40], cos B = tan a cot c. 
log cos c = 9.-75686 
log sec a = 0.03361 
log cos b = 9.79047 
b = 51° S3 7 , 
log sin a = 9.57828 
log esc c = 0.08579 
log sin A = 9.66407 

A = 27° 28' 38". 
log tan a = 9.61188 
log cot c = 9.84266 
log cos B — 9.45454 

B = 73° 27' 11". 

6. Solve the right triangle, given 
c = 23° 49' 51", a = 14° 16' 35". 

By [38], cos b = cos c sec a. 
By [39], sin A = sin a esc c. 
By [40], cos B = tan a cot c. 
log cos c = 9.96130 
log sec a = 0.01362 
log cos 6 = 9.97492 
6 = 19° 17'. 
log sin a = 9.39199 
log esc c = 0.39358 
log sin A = 9.78557 

A = 37° 36' 49". 
log tan a = 9.40562 
. log cot c = 10.35488 
log cos B= 9.76050 

B = 54° 49' 23". 

7. Solve the right triangle, given 
c = 44° 33' 17", a = 32° 9' 17". 

By [38], cos b = cos c sec a. 
By [39] , sin ^. = sin a esc c. 
By [40], cos 5 = tan a cot c. 



log cos c 
log sec a 
log cos b 
b 
log sin a ; 
log esc c : 
log sic ^1 : 

log tan a ; 
log cot c : 
log cos B : 

5: 



: 9.85283 
; 0.07231 
: 9.92514 
: 32° 41'. 
: 9.72608 
: 0.15391 
: 9.87999 
: 49° 20' 16" 
: 9.79840 
: 10.00675 
: 9.80515 
50° 19' 16" 



8. Solve the right triangle, given 
c = 97° 13' 4", a = 132° 14' 12". 

By [38], cos b = cos c sec a. 

By [39] , sin A = sin a esc c. • 

By [40] , cos B = tan a cot c. 

log cos c = 9.09914 (n) 

log sec a = 0.17250 (n) 

log cos b = 9.27164 

b = 79° 13' 38". 
log sin a = 9.86945 
log esc c = 0.00345 
log sin A = 9.87290 

A = 131° 43' 50". 
log tan a = 10.04196 (n) 
log cot c= 9.10259 (n) 
log cos B= 9.14455 

£ = 81° 58' 53". 

9. Solve the right triangle, given 
a = 77° 21' 50", A = 83° 56' 40". 

By [39], sin c = sin a esc A. 

By [42], sin b = tan a cot A. 

By [41], sin B = sec a cos J.. 
log sin a = 9.98935 
log esc A = 0.00243 
log sin c = 9.99178 



250 



SPHERICAL TRIGONOMETRY. 



c = 78° 53' 20", 
or = 101° 6' 40". 

log tan a = 10.64939 
log cot A = 9.02565 
log sin b = 9.67504 

b = 28° 14' 31", 
or = .151° 45' 29". 

log sec a = 0.66004 
log cos A = 9.02323 
log sin B = 9.68327 

B= 28° 49' 57", 
or =151° 10' 3". 

10. Solve the right triangle, given 
a = 77° 21' 50", A = 40° 40' 40". 

By [39], sin c = sin a esc A. 

But sin A < sin a. 

• .-. sin c> 1, which is impossible. 

11. Solve the right triangle, given 
a = 92° 47' 32", B = 50° 2' 1". 

By [40], tan c = tan a sec B. 

By [42], tan b = sin a tan B. 

By [41], cos A = cos a sin B. 
log tan a = 11.31183 (n) 
log sec B = 0.19223 
log tan c = 11.50406 (n) 
c = 91° 47' 40". 
log sin a = 9.99948 
log tan £ = 10.07670 
log tan b = 10.07618 

b = 49° 59' 58". 
log cos a = 8.68765 (n) 
log sin J5 = 9.88447 
log cos A = 8.57212 (n) 
^4 = 92° 8' 23". 

12. Solve the right triangle, given 
a = 2° 0' 55", 5 = 12° 40'. 

By [40] , tan c = tan a sec 5. 
By [42], tan b = sin a tan #. 
By [41], cos A = cos a sin 5. 



log tan a = 8.54639 
log sec B = 0.01070 
log tan c = 8.55709 

c = 2° 3' 56". 
log sin a = 8.54612 
log tan £ = 9.35170 
log tan b = 7.89782 

b = 0° 27' 10". 
log cos a = 9.99973 
log sin B = 9.34100 
log cos A = 9.34073 

A = 77° 20' 28". 

13. Solve the right triangle, given 
a = 20° 20' 20", B = 38° 10' 10". 

By [40], tan c = tan a sec B. 

By [42], tan b = sin a tan B. 

By [41], cos A = cos a sin B. 
log tan a = 9.56900 
log sec B - 0.10448 
log tan c = 9.67348 

c = 25° 14' 38". 
log sin a = 9.54104 
log tan B = 9.89545 
log tan b = 9.43649 

b = 15° 16' 50". 
log cos a - 9.97204 
log sin B = 9.79098 
log cos A = 9.76302 

A = 54° 35' 17". 

14. Solve the right triangle, given 
a = 54° 30', B = 35° 30'. 

By [40], tan c = tan a sec B. 

By [42], tan b = sin a tan B. 

By [41,] cos A = cos a sin B. 
log tan a = 10.14673 
log sec B = 0.08931 
log tan c = 10.23604 

c = 59° 51' 21". 
log sin a = 9.91069 
log tan B = 9.85327 
log tan b = 9.76396 



TEACHERS EDITION. 



251 






b = 30° 8' 39. " 
log cos a= 9.76395 
log sin B = 9.76395 
log cos A - 9.52790 

A = 70° 17' 35". 

15. Solve the right triangle, given 
c = 69° 25' 11", A = 54° 54' 42". 

By [39], sin a = sin c sin A. 

By [40], tan b = tan c cos A. 

By [43], cot B = cos c tan A. 
log sin c = 9.97136 
log sin A = 9.91289 
log sin a = 9.88425 

a = 50°. 
log tan c = 10.42541 
log cos A = 9.75954 
log tan 6 = 10.18495 

b = 56° 50' 49". 
log cos c = 9.54595 
log tan A = 10.15335 
log cot B= 9.69930 
B = 63° 25' 4". 

16. Solve the right triangle, given 
c = 112° 48 / , A = 56° ir 56". 

By [39], sin a = sin c sin A. 

By [40], tan b = tan c cos -4. 

By [43], cot B — cos c tan ^1. 
log sin c = 9.96467 
log sin A = 9.91958 
log sin a = 9.88425 

a = 50°. 
log tan c = 10.37638 (n) 
log cos A = 9.74532 
log tan 6 = 10.12170 (n) 
b = 127° 4 / 30". 
log cos c = 9.58829 (n) 
log tan ^i = 10.17427 
log cot B= 9.76256 (n) 
B = 120° 3' 50". 



17. Solve the ri 

c = 46° 40' 12", ^. 

By [39], sin a 

By [40], tan 6 

By [43], cot B 

log sin c 

log sin ^4 

log sin a 

a 

log tan c 

log cos ^4 

log tan b 

b 

log cos c 

log tan ^i 

log cot B 

B 



ght triangle, given 
= 37° 46' 9". 
= sin c sin A. 
= tan ccos^l. 
= cos c tan ^1. 
= 9.86178 
= 9.78709 
= 9.64887 
= 26° 27' 24". 
= 10.02533 
= 9.89789 
= 9.92322 
= 39° 57' 42". 
= 9.83645 
= 9.88920 
= 9.72565 
= 62° 0' 4". 



18. Solve the right triangle, given 
c = 118° 40' 1", A = 128° 0' 4". 

By [^9], sin a = sin c sin A. 

By [40], tan b = tan c cos A. 

By [43] , cot B = cos c tan A . 
log sin c = 9.94321 
log sin A = 9.89652 
log sin a = 9.83973 

a = 136° 15' 32". 
log tanc = 10.26222(?i) 
log cos A - 9.78935 (n) 
log tan b = 10.05157 

b = 48° 23' 38". 
log cos c= 9.68098 (n) 
log tan A = 10.10717 (n) 
log cot 5= 9.78815 
£ = 58° 27' 4". 

19. Solve the right triangle, given 
A = 63° 15' 12", B = 135° 33' 39". 

By [41], cos a = cos A esc J5. 
By [41], cos& = cos JBcsc^l. 
By [43] , cos c = cot A cot 5. 



252 



SPHERICAL TRIGONOMETRY. 



log cos A = 9.65326 
log esc B = 0.15480 
log cos a = 9.80806 

a = 50° 0' 4". 
log cos B = 9.85369 (to) 
log esc A = 0.04915 
log cos 6 = 9.90284(h) 
b = 143° 5' 12". 
log cot A= 9.70241 
log cot B = 10.00850 (to) 
log cos c = 9.71091 (to) 
. c = 120° 55' 34". 

20. Solve the right triangle, given 
A = 116° 43' 12", B = 116° 31' 25". 

By [41], cos a = cos ^4 esc B. 

By [41], cos 6 = cos B esc ^4. 

By [43], cos c = cot A cot 2?. 
log cos A = 9.65286 (to) 
log esc B = 0.04830 
log cos a = 9.70116 (to) 
a = 120° 10' 3". 
log cos B = 9.64988 (to) 
log esc A = 0.04904 
log cos b = 9.69892 (to) 

b = 119° 59' 46". 
log cot A = 9.70190 (to) 
log cot B = 9.69818 (to) 
log cos c = 9.40008 

c = 75° 26' 58". 

21. Solve the right triangle, given 
A = 46° 59' 42", B = 57° 59' 17". 

By [41], cos a = cos A esc B. 

By [41], cos b = cos B esc A. 

By [43], cos c = cot A cot jB. 

log cos A = 9.83382 

log esc B = 0.07164 

log cos a = 9.90546 

a = 36° 27 7 . 






log cos 5 = 9.72435 
log esc A = 013591 
log cos 6 = 9.86026 

b = 43° 32' 30". 
log cot A = 9.96973 
log cot B = 9.79599 
log cos c = 9.76572 

c = 54° 20'. 

22. Solve the right triangle, given 
A = 90°, B = 88° 24' 35". 

By [41], cos a = cos A esc B. 
By [41], cos b = cos B esc A. 
By [43], cos c = cot A cot B. 
cos ^1 = 0. 
.*. cos a = 0. 
.-.a = 90°. 
esc A = 1. 
.-. b = B. 
.-. 6 = 88° 24' 35' 
cot A = 0. 
.-. cos c = 0. 
.-. c = 90°. 

23. Define a quadrantal triangle, 
and show how its solution may be 
reduced to that of the right triangle. 

A quadrantal triangle is a tri- 
angle that has one or more of its 
sides equal to a quadrant. 

Let A'B'C be a quadrantal tri- 
angle with side A'B' = 90°, or a 
quadrant. 

Let ABC be its polar triangle. 

Then, since 

A'W +C = 180°, C = 90°. 

Hence, ABC is a right triangle. 

Therefore, all parts of the polar 
triangle may be found by formulas 
for the right triangle. 

The parts of A'B'C' may then be 
found by subtracting proper parts 
of ABC from 180°. 









teachers' edition. 253 

24. Solve the quadrantal triangle whose sides are a = 174° 12' 4Q" 
6=94°8 , 20 ,/ , c = 90°. ' 

Let ,4' ^ C, a', 6' c- represent the corresponding angles and sides 
of the polar triangle. 

Then A , = 5047/^ 

B' = 85° 51' 40", 
C = 90°. 
By Prob. 7, Ex. XXXII, 

tan* 1 C ' = - cos (£' + 4') sec ( £' - A% 
By Prob. 8, Ex. XXXII, 

tan* ib' = tan [| (R + A*) - 45°] tan [45° + -j. (& - A')] 
tanH«'= tan[i(^ + -4') - 45°] tan[45° - £(J3' - A*)]. 

B' + A' = 91° 38' SI 77 . 

B' - A' = 80° 4' 29". 
i(£' + .4') _ 45° = 0° 49' 25.5". 
45° + J(JT - A*) = 85° 2 / 14.5". 

45°-i(J5'-.4')z = 4 57'45.5". 

log cos (J5' + ^4') =8.45864 

log sec (B' - ^4') = _o. 76356 

2 ) 9. 22220 

log tan £ c' ="061110 

i c' = 22° 12' 561". 
C = 44° 25' 53". 
C = 135° 34' 7". 
log tan [i(B' + ^ _ 45°] = 8.15770 
log tan [45° + £(£'- 4')] = 11.06133 
2 )~^21903 
log tan £ 6' = 9.60952 

±&' = 22° 8' 35". 
&' = 44° 17' 10". 
B = 135° 42' 50". 
log tan [i(R + ^') _ 450J = 8>15770 
log tan [45° -i(&- A')] = 8.93867 
2 )7.09637 
log tan i a' = 8.54819 

ia'= 2° 1'25". 
a' = 4° 2 7 50". 
A = 175° 57' 10". 



254 



SPHERICAL TRIGONOMETRY. 



25. Solve the quadrantal triangle 
in which c = 90°, A = 110° 47' 50", 
B = 135° 35' 34". 

Let A', B', C", a', 6', c' represent 
the corresponding angles and sides 
of the polar triangle. 

Then a' = 69° 12' 10". 

V = 44° 24' 26". 
C = 90°. 
By [42], tan A' = tan a' esc 6'. 
By [42], tan B' = tan b' esc a'. 
By [38], cos c' = cos a' cos V. 
log tan a' = 10.42043 
log esc V = 0.15505 
log tan A' =10.57548 

i'= 75° 6' 58". 
a = 104° 53' 2". 
log tan b' = 9.99101 
log esc a' = 0.02926 
log tan 5'= 10.02027 

B' = 46° 20' 12". 
b = 133° 39' 48". 
log cos a' = 9.55031 
log; cos &' = 9.85394 



By [39], 
By [40], 
By [43], 



log cos c' = 9.40425 

c' = 75° 18' 21". 
C = 104° 41' 39". 

26. Given in a spherical triangle 
A, C, and c each equal to 90° ; solve 
the triangle. 

By [39] , sin a = sin c sin A 
= 1x1 = 1. 
.-. a = 90°. 
Then B is the pole of 6, and 5 = b; 
but 5 and 6 are otherwise indeter- 
minate. 

27. Given A = 60°, C = 90°, and 
c = 90° ; solve the triangle. 



sin a = sin c sin A. 
tan 6 = tan c cos ^L. 
cot jB = cos c tan A. 
sin a = sin A. 

.>.a = A = 60°. 
tan 6 = oo x| 

= GO. 

.-. b = 90°. 
cot£ = x V3 
= 0. 
.-. B = 90°. 

28. In a right spherical triangle, 
given A = 42° 24' 9", B = 9° 4' 11" ; 
solve the triangle. 

By [43], cos c = cot A cot 5. 
Now cotJ.>l, 

and cotJB>l. 

.-. cose >1, 
which is impossible. 

Therefore, the triangle is impos- 
sible. 

29. In a right spherical triangle, 
given a = 119° 11', 5 = 126° 54'; 
solve the triangle. 

By [42], tan b = sin a tan B. 
By [40], tan c = tan a sec 5. 
By [41], cos A = cos a sin 5. 
log sin a = 9.94105 
log tan 5 = 10.12446 (n) 
log tan b = 10.06551 (n) 
b = 130° 41' 42". 
log tan a = 10.25298 (n) 
log sec 5 = 0.22154 (n) 
log tan c =10.47452 

c = 71° 27' 43". 

log cos a = 9.68807 (n) 

log sin 5 = 9.90292 

log cos^L = 9.59099 (n) 

^ = 112° 57' 2". 



TEACHERS EDITIOX. 



255 



30. In a right spherical triangle, 
given c = 50°, b = 44° 18' 39"; solve 
the triangle. 

By [38], cos a = cose sec b. 

By [40], cos A = tan b cot c. 

By [39], sin B = sin b esc c. 
log cose =9.80807 
log sec b = 0. 14535 
log cos a = 9.95342 

a = 26° 3' 51". 
log tan b = 9.98955 
log cot c = 9.92381 
log cos .4 = 9.91336 

A = 35°. 
log sin b = 9.84419 
log esc c = 0.11575 
log sin 5 = 9.95994 

B = 65° 46'. - 

31. In a right spherical triangle, 
given A = 156° 20' 30", a = 65° 15' 
45" ; solve the triangle. 

The triangle is impossible, because 
a and A are unlike in kind. 

32. In a right spherical triangle, 
given A = 74° 12' 31", c = 64°28'47"; 
solve the triangle. 

By [39], sin a = sin c sin A. 

By [40], tan b = tan c cos A. 

By [43], cot B = cose tan A. 
log sin c = 9.95542 
log sin ul= 9.98329 
log sin a = 9.93871 

a = 60° 16' 17". 
log tan c = 10.32111 
log cos A- 9.43479 
log tan b= 9.75590 

6 = 29° 41' 4", 



log cos c = 9.63431 
log tan A = 10.54851 
log cot B= 10.18282 

B = 33° 16' 54". 

33. In a right spherical triangle, 
given a=112°42 / 38", JS = 44°28 / 44"; 
solve the triangle. 

By [42], tan b = sin a tan B. 
By [40], tan c = tan a sec B. 
By [41], cos^. = cos a sin B. 

log sin a = 9.96495 
log tan B= 9.99210 
log tan b = 9.95705 

b = 42° 10' 17". 

log tan a = 10.37828 (n) 
log sec B= 0.14660 
log tan c = 10.52488 (n) 
c = 106° 37 r 37". 

log cos a = 9.58667 (n) 
log sin £= 9.84550 
logcos^L= 9.43217 (n) 
A = 105° 41' 39". 

34. In a right spherical triangle, 
given 6=48°12'48", ^ = 108°14'44"; 
solve the triangle. 

By [42], tan a = sin b tan A. 
By [40], tan c = tan b sec A. 
By [41], cos B = cos b sin A. 

log sin b= 9.87253 
log tan A= 10.48192 (n) 
log tan a = 10.35445 (n) 
a = 113° 51' 5". 

log tan b = 10.04882 
log sec A = 0.50433 (n) 
log tan c = 10.55315(h) 
c = 105° 37 7 54". 



256 



SPHERICAL TRIGONOMETRY. 



log cos b = 9.82371 
logsin J. = 9.97760 
logeos B= 9.80131 

B = 50° 44' 19". 

35. In a right spherical triangle, 
given J. = 122°68'47", JB= 104°17 / 55 // ; 
solve the triangle. 

By [41], cos a = cos A esc B. 
By [41], cos b = esc A cos B. 
By [43], cos c = cot A cot B. 

log cos A = 9.73587 (n) 
log esc £= 0.01367 
log cos a = 9.74954 (n) 
a = 124° 10' 37". 

log esc ^L = 0.07631 
log cos B= 9.39266 (n) 
log cos b = 9.46897 (n) 
6 = 107° 7' 22". 



logcot^L= 9.81218 (n) 
log cot # = 9.40632 (n) 
log cos c = 9.21850 

c = 80° 28' 49". 



36. If the legs 
spherical triangle 
that cos a = cot A 



By [38], 
But 



By [42], 
But 



cose 
cos a 
. cos c 
cos 2 a 
cos a 
sin b 
sin a 
sin a 



.-. cos a 
.-. cos a 



a and b of a right 
are equal, show 
= Vcos c. 

= cos a cos b. 
= cos b. 
= cos 2 a. 
= c ose. 
= Vcos c. 
= tan a cot A. 

— sin 6. 

= tan a cot A. 
_ sin a cot A 

cos a 
= cot A. 

— cot J. = Vcosc. 



37. In a right spherical triangle show that 

cos 2 .4 sin 2 c = sin (c -f a) sin (c — a). 



By [39], 



sin A — 
sin 2 A = 
.-. cos 2 .4 = 1 — 



sin a 
sin c 
sin 2 a 
sin 2 c 

sin 2 a 



sin 2 c 
_ sin 2 c — sin 2 a 
sin 2 c 
cos 2 A sin 2 c = sin 2 c — sin 2 a. 
By [4], sin (c -{- a) = sin c cos a + cos c sin a. 

By [8], sin (c — a) = sin c cos a — cos c sin a. 

.*. sin (c + a) sin (c — a) = sin 2 c cos 2 a — cos 2 c sin 2 a 

= sin 2 c(l — sin 2 a) — (1 — sin 2 c) sin 2 a 
= sin 2 c — sin 2 c sin 2 a — sin 2 a + sin 2 c sin 2 a 
= sin 2 c — sin 2 a. 
.•. cos 2 J. sin 2 c = sin (c + a) sin (c — a). 



teachers' edition. 257 



38. In a right spherical triangle show that 

tan a cos c = sin b cot B. 
By [42], 

By [43], 



sin b 


= tan a cot A. 


cot A 


sin b 
tan a 


cos c 


— cot ^4 cot B. 


cot A 


cos c 


cot £ 


cose 


sin b 



cot J5 tan a 
tan a cos c = sin b cot 5. 



39. In a right spherical triangle show that 

sin 2 A — cos 2 B -f sin 2 a sin 2 2?. 

By [41], cos B = cos 6sin^l. 

cos B 

.-. sm ^1 = 

sin 2 ^4 = 



cos 6 
cos 2 B 



cos 2 6 
= cos 2 5 sec 2 b 
By Prob. 2, Ex. V, = cos 2 5(1 + tan 2 b) 

= cos 2 5 + tan 2 6 cos 2 5. 
By [42], sin a = tan b cot 5. 

.*. tan b = sin a tan B. 
.-. sin 2 J. = cos 2 2? + sin 2 a tan 2 5 cos 2 B. 
.-. sin 2 A = cos 2 B + sin 2 a sin 2 B. 



40. In a right spherical triangle show that 

sin (b + c) = 2 cos 2 £ ^4 cos 6 sin c. 

By [4] , sin (b + c) = sin 6 cos c + cos 6 sin c 

/ sin 6 cos c „ \ 

= ( h 1 J cos b sm c 

\ cos b sin c / 

= (tan b cot c -f 1) cos 6 sin c. 

By [40], tan 6 cot c = cos ^4. 

.-. tan b cot c + 1 = cos A + 1 

By [17], =2co&±A. 

.-. sin (6 + c) = 2 cos 2 |- ^4 cos 6 sin c. 



258 



SPHERICAL TRIGONOMETRY. 



41. In a right spherical triangle show that 

sin (c — 6) = 2 sin 2 -J A cos 6 sin c. 

By [8], sin (c — b) = sin c cos 6 — cos c sin 6 

. / H cos c sin b \ 

= sin c cos 6(1 ) 

\ sin c cos 6 / 

= sin c cos b (1 — cot c tan 6). 

By [40], cot c tan 6 = cos A. 

.-. 1 — cot c tan 6=1 — cos A 

By [16], =2 sin 2 * A 

.;. sin (c — 6) = 2 sin 2 * ^1 cosl) sin c. 



42. If, in a right spherical triangle, p denotes the arc of the great 
circle passing through the vertex of the right angle and perpendicular to 
the hypotenuse, m and n, the segments of the hypotenuse made by this 
arc adjacent to the legs a and 6, show that (i) tan 2 a = tan c tan m, 
(ii) sin 2 p = tan m tan n. 




(i) In the triangle BCA, by Napier's Rules, 

a) 



COS B : 

.-. tan a ■ 



tan a cot c. 
cos B 



cote 

In the triangle CBD, by Napier's Rules, 

cos B = tan BD cot BC 

= tan m cot a. 

tanm 
.-. tan a = 






cosB 



(2) 



Multiply (1) by (2), tan 2 a = ^ x — 
FJ v ; J w ' cosjB cote 

= tan m tan c. 

(ii) In the triangle CJ3D, by Napier's Rules, 

sin p = tan m cot .BCD. (3) 

In the triangle CAD t by Napier's Rules, 

sin p = tan n cot DC A . (4) 

Multiply (3) by (4), sin 2 p = tan m tan n cot -BCD cot DC A. 

But JSCD + DC A = 90°. 

.-.cot .BCD x cot DCJ. = 1. 

.-. sin 2 p = tan m tan n. 



TEACHERS' EDITION. 



259 



Exercise XXXY. Page 157. 



1. In an isosceles spherical tri- 
angle, given the base b and the side 
a ; find A the angle at the base, B 
the angle at the vertex, and h the 
altitude. 

Let ABA' be an isosceles triangle, 
A and A' the equal angles, a and a' 
the equal sides. 

Draw h, the arc of a great circle, 

from B _L to A A', meeting A A' in C. 

Then, in the right triangle A'BC, 

b — i b in triangle ABA', 

c — a in triangle ABA', 

B — i Bin triangle ABA'. 

By [40] , cos A = cot a tan -J 6. 

By [39] , sin i B = esc a sin h b. 

By [38], cos h = cos a sec -J- b. 

2. In an equilateral spherical tri- 
angle, given the side a ; find the 
angle A. 

In the equilateral triangle A A' A" 
draw arc ^1 C _L to ^i'^i". 
Then, in the right triangle J.^/C, 
sin \ a — sin a sin -£- J.. 
sin ia 



sin £ J. = 
By [12], = 



sin a 
sin i a 



2 sin £ a cos £ a 
= -J sec | a. 

3. Given the side a of a regular 
spherical polygon of n sides; find 
the angle A of the polygon, the 
distance R from the centre of the 
polygon to one of the vertices, and 
the distance r from the centre to the 
middle point of one of the sides. 



In the regular polygon ABBE 
draw arcs of great circles from the 
vertices A, B, etc., through the 
centre O, and from C to M, the mid- 
dle of one side. 




Then ACE = 



ACM = 



360° 
> 

n 

180° 



CAM = I A, 
AM = i a, 
AC = R, 
MC = r. 

By Napier's Rules, 

180 ° 1 • 1 A 

cos = cos $ a sin ■$• A, 

n 

• r, • 180° 

sin i a = sm R sin , 

n 



sin r = tan -J- a cot 



180° 



Whence, 



180 ° 
smii = sec i a cos > 

n 



sin R = sin £ a esc 
sin r == tan i a cot 



180° 

n 
180° 

n 



260 



SPHERICAL TRIGONOMETRY. 



4. Compute the dihedral angles made by the faces of the five regular 
polyhedrons. 

If a sphere is described about a vertex of the polyhedron as a centre 
with a radius equal to an edge of the polyhedron, the adjacent vertices of 
the polyhedron lie on the surface of the sphere and are the vertices of a 
regular spherical polygon, of which the angles are required. 

If a is the length of a side of this polygon, ie., one of the angles of a 
face of the polyhedron, and n the number of sides, ie., the number of 
faces of the polyhedron which meet at a vertex, we have for the different 
cases : 



Polyhedron. 


a 


n 


Tetrahedron . . . 
Hexahedron . . . 
Octahedron .... 
Dodecahedron . . . 
Icosahedron . . . 


60° 

90° 
60° 

108° 
60° 


3 
3 
4 
3 

5 



But if A is an angle of the spherical polygon, we have, from Prob. 3, 

■ 1 a i 180 ° 
sin \ A — sec i a cos 

Hence, for the different cases : 



Polyhedron. 


SinM- 


Log Sin J A. 


A. 


Tetrahedron . . . 


iV3 


9.76144 


70° 31' 46" 


Hexahedron . . . 


W2 


9.84949 


90° 


Octahedron .... 


iV6 


9.91195 


109° 28' 14" 


Dodecahedron . . . 


i sec 54° 


9.92975 


116° 33' 45" 


Icosahedron . . . 


f V3 cos 36° 


9.97043 


138° IV 36" 



5. A spherical square is a regular 
spherical quadrilateral. Find the 
angle A of the square, having given 
the side a. 

This is a special case of Prob. 3 

for which n = 4. 

■ i a i 180 ° 
.*. sin \ A = sec \ a cos 

= \ V2 sec \ a. 



Also cos i A = Vl - i sec 2 £ a. 

cos-M 

.-. cotiA = — 

sin \A 



4 



1 — i sec 2 1 a 
i sec 2 i a 



: V2cos 2 £a- 1 



By [13], = Vcosa. 



TEACHERS 7 edition. 



261 



Exercise XXXVI. Page 160. 



1. What do Formulas [44] be- 
come if A = 90° ? if B = 90° ? if 
C = 90°? if a = 90°? if A = B = 
90°? if a = 6 = 90°? 

If A = 90°, 

sin asm B = sin 6, 
sin a sin C = sin c. 

IiB = 90°, 

sin a = sin b sin .4, 
sin b sin C = sin c. 

If C = 90°, 

sin a = sin c sin J., 
sin 6 = sin c sin 22. 

If a = 90°, 

sin B = sin 6 sin A, 
sin C = sin c sin ^4. 

UA = B = 90°, 

sin a = sin 6, 
sin c = sin a sin C 
= sin b sin C. 

Ka = 6= 90°, 

sin B — sin A, 
sin C = sin c sin ^1 
= sin c sin J5. 



2. What do Formulas [45] be- 
come if A = 90° ? if 5 = 90° ? if 
C = 90° ? if 4 = 5 = C = 90° ? 

If A = 90°, 

cos a = cos b cos c. 

li B = 90°, 

cos 6 = cojs a cos c. 

K C = 90°, 

cos c = cos a cos 6. 

JfA = B=C= 90°, 

cos a = cos b cos c, 
cos b = cos a cos c, 
cos c = cos a cos 6. 

3. What does the first of For- 
mulas [45] become if A = 0° ? if 
^. = 90° ? if J. = 180° ? 

If J. = 0°, 

cos a — cos b cos c + sin b sin c 
= cos (b — c). 
UA = 90°, 

cos a — cos b cos c. 
If JL = 180°, 

cos a = cos b cos c — sin b sin c 
= cos (b + c). 



4. From Formulas [45] deduce Formulas [46] by means of the 
relations between polar triangles (Theorem 4, p. 141). 

Substituting in Formulas [45] for a, 6, c, and A, their equals, 180° — A', 
180° - B', 180° - <7 7 , and 180° - a', we obtain 

cos (180° - A') = cos (180° - B') cos (180° - C) 

+ sin (180° - B') sin (180° - C") cos (180° - a 7 ). 
.-. — cos ^4 7 = cos B' cos C 7 — sin B' sin C 7 cos a'. 
cos ^4/ = — cos B' cos C 7 + sin B' sin C 7 cos a! ; 
and similarly, cos B' = — cos J/ cos C" + sin ^4 7 sin C 7 cos fr 7 ; 
cos C 7 = — cos ^4 7 cos I? 7 + sin ^l 7 sin B' cos c'. 



262 



SPHERICAL TRIGONOMETRY. 



Exercise XXXVII. Page 167. 



1. What are the formulas for 
computing the side a when 6, c, 
and A are given ; and for comput- 
ing the side b when a, c, and B are 
given ? 

(i) In Fig. 100 suppose p drawn 
from C, dividing c into m and n. 

Then the required formulas are 
obtained by advancing the" letters in 
tan m = tan a cos C, 
cose = cos a sec mcos(b — m). 

Hence, the required formulas are 
tan m = tan b cos A , 

cos a = cos b sec m cos (c — m). 

(ii) By drawing p from vi, and 
advancing the letters two steps, 
tan m = tan c cos E, 
cos 6 = cos c sec m cos (a— m). 



2. Given 
a = 88° 12' 20", 
6 = 124° 7' 17", 
C= 50° 2 / 1"; 

*"(&-a) 

*0 

log cos £ (6 — a) 

log sec i (b + a) 

log cot -J- C 

log tan J (5+ 4) 

log sec i {B + A) 

log cos -J- (b + a) 

log sin £ (7 

log cos -| c 



find 

A= 63° 15' 11" 
B = 132° 17' 58" 
c= 59° 4' 17" 

= 17° 57' 28.5". 

= 106° 9' 48.5". 

= 25° V 0.5". 

= 9.97831 

= 0.55536(n) 

= 10.33100 

= 10.86467 (n) 

= 97° 46' 34.7". 

= 0.86868 (n) 

= 9.44464 (n) 

= 9.62622 

= 9.93954 

= 29° 32' 8.6". 



log sin i(b-a) = 9.48900 

log esc i (6 + a) = 0.01751 

log cot £ = 0.33100 

log tan i(B-A)=: 9.83751 

i(B-A)= 34°31'2&.6". 
i (B + A) = 97° 46' 34.7". 

A = 63° 15' 11". 

B = 132° 17' 58". 

c= 59° 4' 17". 

3. Given find 

a = 120° 55' 35", A = 129° 68' 2" 



6= 88° 12' 20" 
C= 47° 42' 1" 



B= 63° 15' 8" 
c= 55° 52' 40" 



i(a-6)= 16° 21' 37.5" 

i(a + &) = 104°33 / 57.5 // 

i(7= 23° 51' 0.5" 



log cos £ (a — b) ■■ 

log sec i (a + 6) : 

log cot £ (7 : 

log tan £ ( J. + B) : 

i(^i + £): 

log sin J (a — 6) : 
log esc i (a + 6) : 

log COt £ C : 

log tan £ (J. - B) : 

A : 

log sec i(A + 5): 

log cos i (a + 6) : 

log sin -J (7 : 

log cos i c -- 

C : 



-• 9.98205 
: 0.59947 (n) 
: 10.35448 
: 10.93600 (n) 

: 96° 36' 35.5". 

: 9.44976 
: 0.01419 
: 10.35448 
: 9.81843 

: 33° 21' 26.7" 
: 96° 36' 35.5" 
: 129° 58' 2". 
: 63° 15' 8". 

• 0.93890 (n) 
: 9.40053 (n) 

9.60675 

9.94618 

: 27° 56' 20". 
: 55° 52' 40". 



TEACHERS EDITION. 



263 



4. Given 

b = 63° 15' 12", 

c = 47°42' 1", 

.4 = 59° 4 / 25 // ; 

*(6 + c) 

l(6-c) 

M 

log cos I (b — c) 

log sec $(b + c) 

log cot -£ A 

log tan i (5 + C) 

i(B+C) 

log sin i (b — c) 

log esc i(b + c) 

log cot -J- A 

log tan i (5- C) 

1(5 -C) 

£(£+C) 

B 

C 

log cos £ (b -f c) 

logsecl(B+ C) 

log sin 1 ^4 

log cos i a 

£a 

a 



find 
B = 88° 12 / 24" 
C = 55° 52' 42" 
a = 50° 1'40" 

= 55° 28' 36.5". 

= 7° 46' 35.5". 

= 29° 32' 12. 5". 

= 9.99599 

= 0.24662 

= 10.24671 

= 10.48932 

= 72° 2' 32.7". 
= 9.13133 
= 0.08413 
= 10.24671 
= 9.46217 
= 16° 9' 51.1". 
= 72° 2' 32.7". 
= 88° 12' 24". 
= 55° 52' 42". 
= 9.75338 
= 0.51101 
= 9.69284 
= 9.95723 
= 25° 0' 50". 
= 50° I 7 40". 



5. Given find 

6 = 69° 25' 11", I? = 56° 11' 57" 
c = 109° 46' 19", C = 123° 21' 12" 
^4 = 54° 54' 42" ; a = 67° ll r 47" 

i (c - b) = 20° 10' 34". 

i (c + 6) = 89° 35' 45". 

i^i = 27° 27' 21". 

logcosi(c-6) = 9.97250 

log sec £ (c + 6) = 2. 15157 

log cot £ ^4 = 10.28434 

log tan i (C + B) = 12.40841 

£(C + JB) = 89° 46' 34.6". 

log sin i (c - b) = 9.53770 

logcsc£(c + b) = 0.00001 

log cot £^1 = 10.28434 

log tan £(<?-£) = 9.82205 

±(C-B)= 33°34 / 37.8" 
C = 123° 2V 12". 
B = 56° 11' 57". 



log cos i (c 4- b) : 

logseci(C + £): 
log sin £ J. : 
log cos i a ■ 



7.84843 

: 2.40842 

9.66376 

9.92061 

: 33° 35' 53.3" 
: 67° 11' 47". 



Exercise XXXVIII. Page 169. 



1. What are the formulas for com- 
puting A when B, C, and a are 
given ; and for computing B when 
A, C, and b are given ? 

(i) In Fig. 101 suppose p drawn 
from C. Then advance the letters 
in 

cot x = tan A esc c, 

cos C = cos A esc x sin (B — x). 



The required formulas are 

cotx =tanjBcsca, 

cos A = cos B esc x sin (C 



•X). 



(ii) Suppose p drawn from A, 
and advance the letters two steps. 
The required formulas are 

cot x = tan C esc b, 

cos B = cos C esc x sin (A — x). 



264 



SPHERICAL TRIGONOMETRY. 



2. Given find 

A = 26° 58' 46", a= 37° 14' 10" 
B = 39° 45' 10", b = 121° 28' 10" 
c = 154° 46' 48"; C = 161° 22' 11" 
i(B-A)= 6°23 / 12". 
i(J5 + ^) = 33°2r S8 77 . 
1 c = 77° 23' 24". 



log cos i(B — A) 

logseci(J5 + ^4) : 

log tan i c ■■ 

log tan i (6 + a) : 

*(6 + a): 

logsini(JB-^): 

logcsci(JB + ^L): 

log tan £ C : 

log tan i(b — a) ■■ 

i(b-a). 

i(b + a): 

b: 



a = 

logsini(J3 + ^L) = 
log sec -J- (b — a) = 
log cos -J- c = 
log cos i C ■ 

iC: 

C 

3. Given 
^4 = 128° 41' 49", 
B = 107° 33' 20", 
c = 124° 12' 31": 



: 9.99730 

: 0.07823 

10.65032 

10.72585 

79° 21' 10.3". 

9.04625 

0.25965 

10.65032 

9.95622 

: 42° r. 

79° 21' 10.3" 
: 121° 28' 10". 
37° 14' 10". 

9.74035 

0.12972 

9.33908 

9.20915 
: 80°4r5.4". 
: 161° 22' 11". 



i(A-B) 

i(A + B) 

i.e 

log cos $(A — B) 

log sec i (^4 + B) 

log tan -J- c 

log tan \ (a + b) 

i(a + b) 



find 

a = 125° 41' 43", 

b= 82° 47' 34", 
C = 127° 22'. 

= 10° 34' 14.5". 

= 118° V 34.5". 

= 62° 6' 15.5". 

= 9.99257 
= 0.32660(71) 
= 10.27624 
= 10.59541 (n) 
= 104° 14' 38.5". 



log sin i ( A — B) : 

log esc i(A + B): 

log tan \ c -. 

log tan i(a — b) ■- 

i(a-b)-. 

a : 

b : 

log sin £(^1 + 5): 

log sec $ (a — b) ■- 
log cos | c : 
log cos i C - 

C: 

4. Given 

£ = 153° 17' 6", 
C= 78°43 / 36", 
a = 86° 15' 15"; 

i(B+C) 

i(B-C) 

ia 

log cos \ (B - O) 

logseci(i?+ C) 

log tan \ a 

log tan i (6 + c) 

*<*+'<?) 

log sin i (J5 - C) 

log esc i (5+ C) 

log tan -J a 

log tan |(6 - c) 

*(&-«) 

6 

c 

log sin i (5+ C) 

log sec i(b — e) 

log cos -J a 

log cos i A 

iA 

A 



-. 9.26351 
: 0.05457 
: 10.27624 
: 9.59432 

: 21° 27' 4.9" 
: 125° 4r 43". 

: 82° 47' 34". 

: 9.94543 
: 0.03118 
: 9.67012 
; 9.64673 
: 63° 41'. 
: 127° 22'. 

find 
b = 152°43 / 51", 
c= 88°12 / 21", 

A= 78°15 / 48". 

= 116° 0' 21". 

= 37°16 / 45". 

= 43° 7' 37-5". 

= 9.90074 
= 0.35807 (n) 
= 9.97159 
= 10.23040(h) 
= 120° 28' 6.2". 

= 9.78226 
= 0.04636 
= 9.97159 
= 9.80021 
= 32°15 / 45". 
= 152° 43 / 51". 
= 88°12 , 21 ,/ . 
= 9.95364 
= 0.07283 
= 9.86322 
= 9.88969 
= 39° 7' 54" 
= 78° 15' 48" 






TEACHERS' EDITION. 



265 



5. Given find 

A = 125° 41' 44", a = 128° 41' 46", 
C = 82° 47' 35", c = 107° 33' 20", 
6 = 52° 37' 57"; J5 = 55° 47' 40" 

£(^4 + C) = 104° 14' 39.5". 

i(A- C) = 21° 27' 4.5". 

f& = 26° IS' 58.5". 

log cos i(JL - C) = 9.96883 
log sec i (A + C) = 0.60896 (n) 
log tan i b = 9.69424 
log tan i (a + c) = 10.27203 (n) 

i(a + c) = 118° 7' 32.9". 



log 


sin J(j4 - 


-C) = 


9.56313 


log< 


2sci(4 + C) = 


0.01356 




log tan ib = 


9.69424 


log 


tan | (a 


-«) = 


9.27093 




i(a 


-c) = 


10° 34' 12.9 






a = 


128° 41' 46". 






c = 


107° 33' 20". 


log 


iini(A + C) = 


9.98644 


log sec J- (a 


-c) = 


0.00743 




log cos i b = 


9.95248 




log cos 


l# = 


9.94635 






iB = 


27° 53' 50". 






J5 = 


55° 47' 40". 



Exercise XXXIX. Page 171. 



1. Given find 

a = 73° 49' 38", B = 116° 42' 30", 
6 = 120° 53' 35", c = 120° 57' 27", 
4= 88°52 / 42"; C = 116° 47'. 

log sin A = 9.99992 
log sin b =9.93355 
log esc a = 0.01753 
log sin B =9.95100 

B = [180° - (63° 17' 30")] 
= 116° 42' 30". 

(The greater side is opposite the 
greater angle.) 

l(J5 + ^) = 102°47 / 36". 
i(B-A)= 13°54 / 54". 
1(5 + a) = 97° 21' 36.5". 

i(b-a)= 23° 31' 58.5". 

log sin i (B + A)= 9.98908 

log esc i (B-A)= 0.61892 

log tan 1 (b - a) = 9.63898 

logtanlc*= 10.24698 

ie= 60° 28' 43.4". 

c = 120° 57' 27". 



log sin | (6 + a) = 9.99641 

log esc i (6- a) = 0.39873 

log tan i(B-A) = 9.39402 





log cot \ C 


= 9.78916 








$C 


= 


58° 23' 30 


". 






c 


= 116° 47'. 




2. 


Given 






find 




a = 


150° 57' 


5", 


B t 


= 120° 47' 


45", 


b = 


134° 15' 


54", 


Cl 


= 55° 42' 


8", 


A = 


144° 22' 


42"; 


Ci 


= 97° 42' 


55"; 








c 2 


= 23° 57' 


17", 








#2 


= 59° 12' 


15", 








c 2 


= 29° 8' 


39". 



A > 90°, (a + b) > 180°, a > b. 
.*. two solutions. 

log sin A- 9.76524 
log sin b =9.85498 
log esc a = 0.31377 
log sin 5 = 9.93399 

B x = 120° 47' 45", 
B 2 = 59° 12' 15". 



266 



SPHERICAL TRIGONOMETRY. 



i(A + B 1 ): 
i(A + B 2 ): 
i(A-B X ): 
i(A-B 2 ): 

i(a-b)-. 
i(a + b): 

log sin i (a + b) ■• 

log esc i(a — b) ■ 

log tan i{A — Bi) ■ 

log cot \ C\ ■ 

i0 1 -. 
-Ox-. 



. 132° 35' 13.5" 
: 101° 47' 28.5" 
: 11° 47' 28.5" 
: 42° 35' 13.5" 
: 8° 20' 35.5" 
142° 36' 29.5" 

: 9.78338 

0.83833 

9.31963 
; 9.94134 

: 48° 5r 27. 7". 
: 97° 42' 55". 



log sin i(a + b) = 9. 78338 

log esc i (a -b) = 0.83833 

log tan i (A - B 2 ) = 9.96338 

logcotiC 2 = 10.58509 



iC 2 : 

C 2 : 

logsini(^ + ^i): 
log csc \ (A — Bi) ■ 
log tan i (a — b) ■ 
log tan i Ci 

ci 

log sin \ {A + B 2 ) : 

log csc i(A — B 2 ) : 

log tan i(a — b) ■■ 

log tan i c 2 ■ 

%c 2 . 

C 2 : 



: 14° 34' 19.6" 
:29° 8' 39". 

: 9.86703 
: 0.68963 
: 9.16629 
: 9.72295 

: 27° 51' 4". 

: 55° 42' 8". 

: 9.99074 
: 0.16960 
= 9.16629 
: 9.32663 

: 11° 58' 38.7" 
: 23° 57' 17". 



3. Given 

a = 79° 0'54", 

b = 82° 17' 4", 

A = 82° 9' 26"; 

log sin A 

log sin b 

colog sin a 

log sin B 

B 

tan c 

cotC 

log cos A 

log tan b 

log tan c 

c 

log tan A 

log cos b 

log cot O 

C 



find 
B = 90°, 
c = 45° 12' 19" 
C = 45°44' 5" 

= 9.99592 

= 9.99605 

= 0.00803 

= 0.00000 

= 90°. 

= cos A tan b. 

= tan A cos 6. 

= 9.13499 

= 10.86812 

= 10.00311 

= 45° 12' 19". 

= 0.86093 

= 9.12793 

= 9.98886 

= 45° 44' 5". 



4. Given a = 30° 52' 37", b = 
31° 9' 16", A = 87° 34' 12" ; show 
that the triangle is impossible. 
By [44], sin B = sin A sin b csc a. 
log sin A = 9.99961 
log sin b =9.71378 
log csc a = 0.28972 
log sin B =0.00311 
sin B = 1.0072. 
Therefore, the triangle is impos- 
sible, since sin B > 1. 



1. Given 
A = 110° 10', 
133° 18', 



B 



Exercise XL. Page 173. 



find 
b = 155° 5' 18' 
c = 33° 1'37 A 



a = 147° 5' 32"; C = 70° 20' 40" 



log sin a =9.73503 
log sin £ = 9.86200 
log csc ^4 = 0.02748 
log sin b = 9.62451 



TEACHERS' EDITION. 



267 



i(B + A) 

i(B-A) 
i(b-a): 
i(6 + a): 

log sin i( A) -. 

log esc i(B — A) : 

log tan i (b — a) ■ 

log tan } c = 9.47198 

\c = 16° 30' 48.5" 

C : 



155° 5' 18" 
: 121° 44'. 
: 11° 34'. 
: 3° 59' 53" 
: 151° 5' 25" 

9.92968 
: 0.69787 

8.84443 



log esc i (b — a) • 

log sin i (b + a) ■ 

log tan i (5- A) : 

log cot i C : 

C 

2. Given 

^i = 113° 39' 21", 

B = 123° 40' 18", 

a= 65° 39' 46"; 

log sin a 

log sin B 

log esc A 

log sin 6 

6 

i(B + A) 

i(B-A) 

i(b-a) 

i(b + a) 

log sin i (B + A) 

log esc \(B — A) 

log tan ^ (6 — a) 

log tan -J- c 

*c 

c 

log sin i (b + a) 

log esc i(b — a) 

I logtani(B--4) 

i , log cot i C 



33° 1'37". 
; 0.15663 

9.68433 

9.31104 

9.15200 

35° 10' 20". 
: 70° 20' 40". 

find 
b = 124° 7' 20" 
c = 159° 50' 15" 
C = 159° 43' 34" 

= 9.95959 

= 9.92024 

= 0.03812 

= 9.91795 

= 124° 7' 20". 

= 118° 39' 49.5". 

= 5° 0'28.5". 

= 29° 13' 47". 

= 94° 53' 33". 

= 9.94322 

= 1.05902 

= 9.74785 

= 10.75009 

= 79° 55' 7.3". 

= 159° 50' 15". 

= 9.99841 

= 0.31130 

= 8.94264 

= 9.25235 



iC= 79° 51' 46.8" 
C = 159° 43' 34". 



3. Given 
A = 10C 2' 11" 
B = 98° 30' 28" 
a= 95° 20' 39". 

log sin a -. 
log sin B ■■ 
log esc A : 
log sin b ■■ 
b-. 

i(A+B): 
i(A-B): 

i(a-b). 

Ha + 6): 

log sin i(A + B): 

log esc i(A — B) • 

log tan i(a — b) -. 

log tan £ c ■■ 

C : 

log sin i(a + b) ■ 

log esc i (a — b) : 

log tan i (A - B) -. 

log cot i C -- 

iC: 
C: 



find 
b= 90°, 
c = 147° 41' 50" 
C = 148° 5' 40" 

: 9.99811 
9.99519 
0.00670 



10.00000 
90°. 

99° 16' 19.5". 

: 0° 45' 51.5". 

2° 40' 19.5". 

92° 40' 19.5". 

9.99428 
1.87487 
8.66904 



; 10.53819 

: 73° 50' 54.9" 

: 147° 41' 50". 

: 9.99953 

1.33144 

8.12517 

9.45614 
: 74° 2' 50". 
: 148° 5' 40". 



4. Given A = 24° 33' 9", B = 
38° 0' 12", a = 65° 20' 13" ; show 
that the triangle is impossible. 

log sin a = 9.95845 
log sin 5= 9.78937 
log esc A = 0.38140 
log sin b =10.12922 
.*. sin b > 1. 
Therefore, the triangle is impos- 
sible. 



268 



SPHERICAL TRIGONOMETRY. 





Exercise XLL Page 174. 




1. Given find 


s = 148° 4' 17". 


a 


= 120° 55 7 35", A = 116° 44' 50", 


s-a= 97°52 / 13". 


b 


= 59° 4' 25", B = 63° 15' 10", 


s - b = 31° 19' 29". 


c 


= 106°10 7 22"; C= 91° V 22". 


s -c= 18° 52 7 35". 




a - 120° 55' 35" 


log sin (s- a) = 9.99589 




b= 59° 4' 25" 


log sin (s- b) = 9.71591 




c = 106° 10' 22" 


log sin (s - c) = 9.50992 




2s = 286° 10' 22" 


log esc s = 0.27666 




s = 143° 5 7 11". 


log tan 2 r = 9.49838 




s-a= 22° 9' 36". 


log tan r = 9.74919. 




s _6 = 84° 0'46". 


logtani^l= 9.75330 




s _ c = 36° 54 / 49". 


logtan££ = 10.03328 




log sin (s- a) = 9.57657 


log tan tC = 10.23927 




log sin (s -b) = 9.99763 


iA= 29° 32' 14". 




log sin (s - c) = 9.77859 


iB= 47°11 7 36". 




log esc s = 0.22141 


iC= 60° 2 / 26". 




log tan 2 r = 9.57420 


A = 59° 4' 28". 
B = 94° 23' 12". 




log tan r = 9.78710. 


C = 120° 4 7 52". 




logtani^L = 10.21053 






logtan£J5 = 9.78947 


3. Given find 




logtaniC = 10.00851 


a = 131° 35 7 4", A = 132° 14 / 21", 




iA= 58° 22' 24.8". 


b = 108° 30' 14", B = 110° 10 7 40", 




iB= 31°37 / 35.2". 


c= 84° 46' 34"; C = 99° 42' 24". 




iC= 45° 33 7 40.8". 


a = 131° 35' 4" 




A = 116° 44 7 50". 


6 = 108° 30 7 14" 




B = 63°15 7 10". 


c= 84°46 / 34" 




C= 91° 7 / 22". 


2 s = 324° 51' 52" 
s = 162° 25 7 56". 




2. Given find 


s-a= 30° 50' 52". 


a 


= 50°12 / 4", A= 59° 4' 28", 


s - b = 53° 55 7 42". 


b 


= 116°44 7 48", 5= 94°23 7 12", 


s - c = 77° 39 7 22". 


c 


= 129°ir42"; O=120° 4' 52". 


log sin (s- a) = 9.70991 




a= 50° 12' 4" 


log sin (s - 6) = 9.90756 




6 = 116° 44' 48" 


log sin (s-c)= 9.98984 




c = 129° ll 7 42" 


log esc s= 0.52023 




2s = 296° 8' 34" 


log tan 2 r = 10.12754 



teachers' edition. 



269 



log tan r = 10.06377. 
log tan ^z= 10.35386 


s = 71° 28' 31". 
s- a = 51° 11' 53". 


logtaniJ5= 10.15621 


s-b= 15° 8' 51". 


logtan£C = 10.07393 


8 - c = 5° 7' 47". 


iA= 66° 7' 10.6". 


log sin (s -a) = 9.89172 


iB= 55° 5' 20". 


log sin (s- b) = 9.41715 


i C = 49° 51' 12". 


log sin {s~c) = 8.95139 


A = 132° 14' 21". 


log esc s = 0.02311 


B = 110° 10' 40". 


log tan 2 r = 8.28337 


Cj= 99° 42' 24". 


log tan r = 9.14169. 


4. Given find 
a = 20° 16' 38", A= 20° 9' 55" 
b = 56° 19' 40", 5= 55° 52' 35",' 
c = 66° 20' 44" ; C = 114° 20' 21". 
a = 20° 16' 38" 
b= 56° 19' 40" 
c = 66° 20' 44" 
2 s = 142° 57' 2" 


logtani^l = 9.24997 
log tan i B= 9.72454 
logtan|C = 10.19030 

iA= 10° 4' 57.6" 

iB= 27° 56' 17.4" 

iC= 57° 10' 10.7". 

A = 20° 9' 55". 

B= 55° 52' 35". 

C= 114° 20' 21". 


Exercise XLI 


I. Page 176. 


1. Given find 


logtan£a = 10.43139 


A = 130°, a = 139° 21' 22", 


logtan|6 = 10.30193 


B = 110°, 6 = 126° 57' 52", 


logtan£c= 9.73353 


C = 80° ; c = 56° 51' 48". 

A = 130° 

B = 110° 

C = 80° 

2S = 320° 


ia= 69° 40' 41". 
ib= 63° 28' 56.2". 
ic= 28° 25' 54". 

o = 139° 21' 22". 

b = 126° 57' 52". 


£ = 160°. 


c = 56° 51' 48". 


S - A = 30°. 




S~B = 50°. 


2. Given find 


S- C = 80°. 


A = 59° 55' 10", a = 51° 17' 31", 


log cos 5= 9.97299 (n) 


B = 85° 36' 50", b = 64° 2' 47", 


log sec (S-4)"= 0.06247 


C = 59° 55' 10" ; c = 51° 17' 31".' 


log sec (S - B) = 0.19193 


A = 59° bb f 10" 


log sec (S- O = 0.76033 
log tan* £ = 10.98772 
logtani? = 10.49386. 


B = 85° 36' 50" 

C = 59° 55' 10" 

2S = 205° 27' 10" 



270 



SPHERICAL TRIGONOMETRY. 



S = 102° 43' 35" 

S-A = 42° 48' 25" 
S- B = 17° 6' 45" 

5 - C = 42° 48' 25" 



log cos S 

log sec (S - ^4.) 

log sec (S - B) 

log sec (S - C) 

log tan 2 i? 

log tan B 

log tan £ a : 
log tan i b ■- 
log tan ic ■ 

£a: 

iC: 

a : 

b: 
C : 



= 9.34301 (ft) 
= 0.13451 
= 0.01967 
= 0.13451 
= 9.63170 

= 9.81585. 

= 9.68134 
= 9.79618 
= 9.68134 

: 25° 38' 45.5" 
:32° 1'23.6" 
:25°38 / 45.5" 
: 51° 17' 31". 
:64° 2' 47". 
: 51° 17' 31". 



3. Given find 

A = 102° 14' 12", a - 104° 25' 9" 

B= 54° 32' 24", b= 53° 49' 25" 

C= 89° 5' 46"; c= 97° 44' 19" 

A = 102° 14' 12" 
B= 54° 32' 24" 

C= 89° 5' 46" 



2S = 245° 52' 22" 



S: 
8-A: 
S-B: 

S-C: 

log COS S : 

log sec (S — A) ■. 

log sec {S-B)-. 

log sec (8 - C) ■■ 

log tan 2 12: 

log tan B ■ 



122° 56' 11" 
20° 41' 59" 
68° 23' 47" 
33° 50' 25" 

9.73536 (n) 
0.02898 
0.43394 
0.08061 



10.27889 
10.13945. 



log tan i a = 10.11047 
logtan*fe= 9.70551 
log tan £ c = 10.05884 

ia= 52° 12' 34.6" 
ib= 26° 54' 42.6" 
ic= 48° 52' 9.6". 

a = 104° 25' 9". 

b= 53° 49' 25". 

c = 97° 44' 19". 



4. Given 

A= 4° 23' 35", 

B = 8° 28' 20", 
C = 172° 17' 56"; 

^1 = 
B = 
C = 

2S = 

8 = 
S-A = 
S-B = 

s- c = - 

log cos S ■ 

log sec (S — A) -- 

log sec (S — -B) : 

log sec (S-C)-. 

log tan 2 B-. 



find 

a= 31° 9' 13", 
b= 84° 18' 28", 
c= 115° 10'. 

4° 23' 35" 

8° 28' 20" 

172° 17' 56" 

185° 9' 51" 

92° 34' 55.5". 
88° 11' 20.5". 
84° 6' 35.5". 
(79° 43' 0.5"). 

: 8.65370 (n) 

: 1.50029 

: 0.98876 

: 0.74833 



11.89108 
log tan 12 = 10.94554. 



log tan -j- a ■ 
log tan £ b : 
log tan \ c ■. 

ia : 
ib 

a 



9.44525 

9.95678 

10.19721 

15° 34' 36.7" 

42° 9' 13.8" 

57° 35'. 

31° 9' 13". 
' 84° 18' 28". 
115° 10'. 



TEACHERS EDITION. 



271 



Exercise XLIII. Page 180. 



1. Given find 

A = 84° 20' 19", E = 26159", 
B = 27° 22' 40", F = 0, 12682 R 2 . 
C = 75° 33'; 

A = 84° 20' 19" 
B= 27°22 , 40" 
C= 75° 33' 
i + 5+C = 187° 15' 59" 

.-. # = 7° 15' 59" 
= 26159". 
logtf^logE 2 
log 26159 = 4.41762 

log — — = 4.68557- 10 
6 648000 



log F = 9. 10319 - 10 + log i? 2 
F = 0.12682 £*. 



2. Given 




find 


a= 69° 15' 6 


", E 


= 216° 40' 18 


b = 120° 42' 47 


", 




c = 159° 18 / 33 


> 






a = 


69° 15' 6" 




6 = 


120° 42' 47" 




c = 


159° 18' 33" 




2s = 


349° 16' 26" 




s = 


174° 38' 13". 


5 - 


- a = 


105° 23' 7". 


S - 


-6 = 


53° 55' 26". 


S - 


- c = 


15° 19' 40". 


is 


= 


87° 19' 6.5" 


i(s- 


a) = 


52° 41' 33.5" 


i(s- 


&) = 


26° 57' 43". 


i(s- 


c) = 


7° 39' 50". 


log tan -j- s = 


11.32942 


log tan i (s — 


a) = 


10.11805 


log tan i (s — 


&) = 


9.70645 


log tan i (s — 


c) = 


9.12893 


log tan 2 % 


^ = 


10.28285 





log tan i E 


= 10.14142. 






iE 


= 54° 10' 4.6" 






E 


= 216° 40' 18". 


3. 


Given 




find 


a = 


33° 1° 


45", 


E = 133° 48' 53 


b = 


155° 5' 


18", 




C = 


110° 10' 







By Sect. LVII, 

tan m = tan a cos C, 
cos c = cos a sec m cos (b — m) 

log tan a = 9.81300 
log cos C — 9.53751 
log tan m = 9.35051 

m = 167° 22'. 
b - m =-(12° 16' 42"). 

log cos a = 9.92345 
log sec ra = 0.01064 (n) 
log cos (b - m) = 9.98995 

log cos c = 9.92404 (n) 

c = 147° 5' 30". 

a-= 33° 1'45" 
b = 155° 5' 18" 
c = 147° 5' 30" 



2 s = 335° 12' 33" 



s = 


167° 36' 16.5". 


s — a = 


134° 34' 31.5". 


s-6 = 


12° 30' 58.5". 


s - c = 


20° 30' 46.5". 


is = 


83° 48' 8.25" 


i(s-a) = 


67° 17' 15.75" 


±(s-b) = 


6° 15' 29.25" 


}(s-c) = 


10° 15' 23.25" 



272 



SPHERICAL TRIGONOMETRY. 



logtan|s = 10.96419 


i(6-a) = 33°13'. 


log tan i(s - a) = 10.37824 

log tan i(s- b) = 9.04005 

logtan£(s- c) = 9.25755 

logtan 2 iJ£ = 9.64003 


log sin £ (£ + J.) = 9.98879 

log sec i (6- a) = 0.07748 

log cos ic =9.73338 

log cosi (7 = 9.79965 


log tan IE= 9.82002. 


iC= 50° 55'. 


iE= 33° 27' 13.3". 


C = 101° 50'. 


E = 133° 48' 53". 


A = 78° 42' 33" 




B = 127° 13' 7" 


4. Given c = 114° 27' 57", ^L = 


C = 101° 50' 


78° 42' 33", 5 = 127° 13' 7" ; find 


A + 5 + C = 307° 45' 40" 


the area. 


.-. .E = 127° 45' 40" 


i(B-A)= 24°15 / 17". 


= 459940". 


*(£ + .!) = 102° 57' 50". 


log £ 2 = log jR 2 


£c = 57°13 / 58.5". 


log 459940 = 5.66270 


log sin i(B-A)= 9.61362 


Tt 


logcsci(B + ^.)= 0.01121 


1Og 648000- 4 ' 68557 - 10 


logtanic = 10.19128 


log F= 0.34827 + log E 2 


log tan i (b- a) = 9.81611 


F= 2.2298 £ 2 . 


5. Given a = 76° 14' 47", b = 82 c 


40' 15", -4 = 60° 22' 44" ; find the 


area. 





Here a and A are alike in kind and sin b > sin a > sin ^4 sin &. 
Hence, there are two solutions. 

log sin A = 9.93918 
log sin b =9.99643 
log esc a = 0.01264 
log sin B =9.94825 

B = 62° 34' 51", 
and 5 1 =117°25 / 9". 

i(b + a) = 79° 27' 31". 

i(b-a)= 3° 12' 44". 
i(B + A) =61° 28' 47.5". 
i(B-A)= 1° 6' 3.5". 

log sin 1(5 + JL)= 9.94382 

log esc i (B-A)= 1.71637 

logtan£(6 - a) = 8.74914 

logtan£c = 10.40933 



TEACHERS EDITION. 



273 



ie= 08° 42' 42.6". 
c= 137° 25 / 25 // . 
i(Bt + A)= 88° 53' 56. 5". 
i(Bi-^l) = 28°31 / 12.5 // . 

log sin -j- (i? x + A) = 9.99992 

log esc i (B 1 -A) = 0.32105 

log tan i(b-a) = 8.74914 

logtan£ci = 9.07011 

id= 6°42 / 9.4". 
ci = 13° 24' 19". 



a : 

b: 

C : 

2S: 



: 76° 14' 47" 

: 82°40 / 15 // 

137° 25' 25" 

; 296° 20' 27" 



76° 14' 47" 

82° 40' 15" 

or 13° 24 7 19" 

or 172° 19' 21" 



s = 148° 10' 13.5" or 86° 9' 40.5 
s — a 
s-b 
s — c 
is 
i(s-a) 
i(s-b) 



71° 55' 26.5" 
65° 29' 58.5" 
10° 44' 48.5" 
74° 5' 6.75" 
35° 57' 43.25" 
32° 44' 59.25" 



or 
or 
or 
or 
or 
or 



9°54 / 53.5". 

3° 29' 25.5". 
72°45 / 21.5". 
43° 4' 50.25" 

4° 57' 26.75" 

1° 44' 42.75" 



5° 22' 24.25" or 36° 22' 40.75" 



log tan i s : 

log tan i (s — a) ■ 

log tan i (s — b) ■ 

log tan i(s — c) ■ 

log tan 2 i E ■ 

log tan i E ■ 



10.54494 or 9.97088 
9.86065 or 8.93822 
9.80836 or 8.48386 
8.97340 or 9.86727 
9.18735 or 7.26023 

9.59368 or 8.63012. 



iJ£ , = 21°25 / 22.7" or 2° 26' 36.0" 
E = 85° 41' 31" or 9° 46' 24". 
^ = 308491" or 35184". 



log 



log R 2 : 

\OgE - 

It 
648000 " 
logF: 



: 85° 41' 31' 

: 308491" 

: log R 2 

-. 5.48925 or 

: 4.68557 - 10 



logR 2 
4.54635 

4.68557 - 10 



F = 



0.17482 + logR 2 or 9.23192 - 10 + log R 2 
1.4956 R 2 or 0.17085 E 2 . 



274 



SPHERICAL TRIGONOMETRY. 



6. Given A = 80° 12' 35", B = 
77° 38' 22", a = 76° 42' 28" j find 
the area. 
By Sect. LX, Note 2, 

cot x = cos a tan 1?, 
sin (C — x) = cos A sec 5 sin x. 
log cos a = 9.36157 
log tan £= 10.65927 
log cot a; = 10.02084 

x = 43° 37 / 34". 
log cos .4 = 9.23055 
log sec B = 0.66946 
log sin x = 9.83881 
logsin(C-x) = 9.73882 
C-x= 33° 14'. 
C= 76° 51' 34". 
A = 80° 12' 35" 
B= 77° 38' 22" 
C= 76° 51' 34" 
A + B + C = 234° 42 / 31" 

.\J£ = 54° 42' 31" 
= 196951". 
log R 2 = log R 2 
log #=5.29436 



log 



it 



648000 



= 4.68557 - 10 



log F = 9.97993 - 10 + log R 2 
F = 0.95484 R 2 . 



7. Given 6 = 44° 27' 40", c = 
15° 22' 44", A = 167° 42' 27" ; find 
the area. 

i(&-c) = 14°32'28". 
i(6 + c) = 29°55'12". 
£.4 = 83° 51' 13.5". 
log cos i (b -c) = 9.98586 
log sec i (b + c) = 0.06212 
log cot £.4 = 9.03215 
log tan i {B + C) = 9.08013 



£(£+ C) = 6° 51' 27.5" 

£ + C = 13° 42' 55" 

A = 167° 42' 27" 

B + C= 13° 42' 55" 

A + 5 + C = 181° 25' 22" 

.-. E = 1° 25' 22" 

= 5122". 

log £ 2 = lOg E 2 

log# = 3.70944 



log 



it 



648000 



= 4.68557-10 



log JP 1 = 8.39501 - 10 + log R 2 
F = 0.024832 R 2 . 



8. Given b = 67° 15' 42", A = 
84° 55' 8", C = 96° 18' 49"; find the 
area. 

By Sect. LVIII, 

cot x = tan C cos 6, 
cos B = cos C esc x sin (A — x). 
log tan C = 10.95608 (w) 
log cos b = 9.58718 
log cot x = 10.54326 (n) 
x = 164° V 35". 
logcosC= 9.04128 (it) 
log csc x = 0.56036 
log sin ( A - x) = 9.99210 (n) 
log cos £ = 9.59374 

£= 66° 53' 44". 
A= 84° 55' 8" 
J5= 66° 53' 44" 
C= 96° 18' 49" 
A + £ + C = 248° 7' 41" 
.-. E = 68° 7' 41" 






= 245261' 
log R 2 = log E 2 
log # = 5.38963 



log 



it 



648000 
logF-- 
F-- 



4.68557 - 10 






: 0.07520 + log R 2 
1.1891E 2 . 



TEACHERS' EDITION. 



275 



9. Given b = 72° 19' 38", c = 
54° 58' 52", B = 77° 15' 14"; find 
the area. 

By Sect. LIX, Note 2, 

tan m = cos B tan c, 
cos (a — m) = cos 6 sec c cos m. 

log cos 5= 9.34367 
log tan c = 10.15447 
log tan m= 9.49814 

m = 17° 28 / 41". 





log( 


zosb = 


9.48227 




log* 


sec c = 


0.24121 




log cos m = 


9.97947 


log cos (a - 


- ?n) = 


9.70295 




a 


— m — 


59° 41' 41" 






a = 


77° 10' 22". 






a = 


77° 10' 22" 






b = 


72° 19' 38" 






C — 


54° 58' 52" 






2s = 


204° 28' 52" 






s = 


102° 14' 26". 




s 


— a — 


25° 4' 4". 




s 


-b = 


29° 54' 48". 




s 


- c = 


47° 15' 34". 






is = 


51° 7' 13". 




i(s- 


-a) = 


12° 32' 2". 




i(s 


-b) = 


14° 57' 24". 




i(s 


-c) = 


23° 37' 47". 




log tan -£- s = 


10.09350 


log 


tan £ (s - 


-a) = 


9.34697 


log 


tan J (s - 


-b) = 


9.42673 


log 


tan^(s - 


-c) = 


9.64099 




log tan 2 


IE = 


8.50819 




log tan 


IE = 


9.25410. 






hE = 


10° 1<K 37.5" 






E = 


40° 42' 30" 






■=. 


146550". 



log 



lOg #2 = log R2 

logF = 5.16599 
it 



= 4.68557 - 10 



648000 
log F = 9.85156 - 10 + log W 



C = 
find 



F = 0.7105E 2 . 

10. Given B = 127° 16' 4 
42° 34' 19", b = 54° 47' 55" 
the area. 

sin c = sin b sin C esc B. 
log sin 6 = 9.91229 
log sin C = 9.83027 
log esc B = 0.09919 
log sin c = 9.84175 

c = 43° 59' 51". 
±(S+ G) = 84° 55' 11.5". 
i(£- C) = 42°20'52.5". 
^ (6 + c) = 49° 23' 53". 
J (6 - c) = 5° 24' 2". 
log sin £ (6 + c) = 9.88039 
log esc i (b - c) = 1.02633 
logtan i (5 - C) = 9.95974 
log cot -M = 10.86646 

iA= 7°44 / 41.1". 
A = 15°29 / 22". 
A = 15° 29' 22" 
B = 127° 16' 4" 
C= 42° 34' 19" 
A -f JS + C = 185° 19' 45" 
.-. E= 5°19 , 45" 
= 19185". 
logE 2 = logE* 2 
log# = 4.28296 

7t 



log 



648000 



: 4.86557 - 10 



logF = 8.96853 - 10 + logi? 2 
F= 0.09301 R 2 . 

11. Given a = 128° 42' 56", b = 
107° 13' 48", c = 88° 37' 51"; find 
the area. 



276 



SPHERICAL TRIGONOMETRY. 



a = 128° 42' 56" 
& = 107° 13' 48" 
c = 88° 37' 51" 
2s = 324° 34' 35" 
8= 162° 17' 17.5". 
s-a = 33° 34' 21.5". 
s- 6 = 55° 3' 29.5". 
s-c= 73° 39' 26.5". 
is= 81° 8' 38. 75". 
i(s-~a)= 16°47 / 10.75". 
i(s-b)= 27° 3r 44. 75". 
i(s- c)= 36° 49' 43.25". 
log tan is = 10.80742 
log tan i (s - a) = 9.47951 
logtani(s- b)= 9.71701 
log tan i(s - c)= 9.87441 
logtan 2 i^= 9.87835 
logtani^= 9.93918. 

IE = 41° 0' 4.6". 
E = 164° 0' 18" 
= 590418". 
log^^rlogi? 2 
log^ = 5.77116 



log 



648000 
logF-- 



= 4.68557 - 10 



: 0.45673 + log B 2 
F= 2.8624 jR 2 . 
12. Given A = 127° 22' 28", B = 
131° 45' 27", C = 100° 52' 16" ; find 
the area. 

A = 127° 22' 28" 
B = 131° 45 / 27" 
C = 100° 52' 16" 
A + jS + C = 360° 0' 11" 
.-.^ = 180° 0' 11" 
= 648011". 
logE 2 = logE 2 
logJ£ = 5.81159 



log 



Tt 



648000 



: 4.68557 - 10 



logjF= 0.49716 4- log R 2 
.F=3.1416E 2 . 



13. Given a = 116° 19' 45", A = 
160° 42' 24", C = 171° 27' 15"; find 
the area. 

By Sect. LX, Note 2, 

cot x = cos a tan C, 

sin (B — x) = cos A sec C sin x. 

log cos a = 9.64692 (n) 

log tan C = 9.17687 (n) 

log cot x = 8.82379 

x = 86° 11' 13". 
log cos A = 9.97490 (n) 
log sec C = 0.00485 (n) 
log sin x = 9.99904 
log sin (B-x) = 9.97879 

B-x= 72° 14' 15". 
B = 158° 25' 28". 
A = 160° 42' 24" 
B = 158° 25' 28" 
C = 171° 27' 15" 
A + 5 + C = 490° 35' 7" 
.-. E = 310° 35' 7" 
= 1118107". 
logE 2 = logE 2 
logJ£ = 6.04848 

= 4.68557 - 10 



it 



648000 



logJP = 0.73405 + log £ 2 
F=5.4206i2 2 . 

14. Find the area of a triangle 
on the earth's surface (regarded as 
spherical) if each side of the triangle 
is equal to 1°. (Radius of earth — 
3958 miles.) 

a = l° 

b = l° 

c = P 

2s = 3° 

s = 1° 30'. 

s - a = 0° 30'. 

s - 6 = 0° 30'. 



TEACHERS' EDITION. 



277 



s - c = 0° 30'. 
|s = 0°45 / . 
£(s-a) = 0°15'. 
l(s - 6) = 0° 15'. 
J(«-c) = 0°15'. 
log tan is = 8.11696 
log tan i(8 - a) = 7.63982 
log tan i(s-b) = 7.63982 
log tan i (s -c) = 7.63982 
logtan 2 ±-E = 1.03642 
logtan±J£= 5.51821. 



log 



iE = 0°0' 6.8139" 
E = 0°0' 27.2556" 
log # 2 = 7.19496 
log # = 1.43546 

= 4.68557 - 10 



648000 



log F= 3.31599 
F = 2070.1. 



Therefore, the area is 2070.1 
square miles. 



Exercise LXIV. Page 197. 



1. Find the dihedral angle made 
by adjacent lateral faces of a regular 
ten-sided pyramid ; given the angle 
V = 18°, made at the vertex by two 
adjacent lateral edges. 

About the vertex of the pyramid 
describe a sphere. The surface of 
the sphere will intersect the lateral 
surface of the pyramid, forming a 
regular spherical decagon, of which 
each side a is equal to 18°, being 
measured by the plane angle at the 
centre of the sphere. 

The required angle is an angle A 
of this decagon. 

By Prob. 3, Ex. XXXV, 

sin i A = sec i a cos 

10 
= sec 9° cos 18°. 
log cos 18° = 9.97821 
colog cos 9° = 0.00538 
log sin i A = 9.98359 
$A= 74° 21'. 
A = 148° 42'. 

2. Through the foot of a rod 
which makes the angle A with a 
plane a straight line is drawn in 
the plane. This line makes the 



180° 



angle B with the projection of the 
rod upon the plane. What angle 
does this line make with the rod ? 

Let CO be a straight line, making 
the angle A with the plane GH ; 
OI a straight line passing through 
the foot of CO, making the angle B 
with the projection EO of CO upon 
the plane GH. 

a 




It is required to find the angle 
COI = x. 

With a radius equal to unity, 
from O as a centre, construct the 
spherical triangle DC I. 
Then i = A, 

c = B, 

d = COI = x, 
and CDI = 90°. 

By [38], cos d = cos i cos c. 
.-. cos x = cos A cos B. 



278 



SPHERICAL TRIGONOMETRY. 



3. Find the volume V of an oblique parallelopipedon ; given the three 
unequal edges a, 6, c, and the three angles Z, m, n, which the edges make 
with one another. 



-4 




c 






">- m 


V^ 


*$■ 




\ ' \ n 
\ a \ 


c \ 


z \ 






,.cc" 


/^u 




D 


c 





Let AB be an oblique parallelopipedon, and Z, m, and ?i, the angles 
which the unequal edges a, 6, and c make with one another. 
Required the volume, V. 
Let w equal the inclination of the edge c to the plane of a and b. 

Now, by Geometry, V = area base x altitude. 

By Prob. 12, Ex. XXI, Area base = ab sin Z. 

Altitude = x = c sin w. 
.-. T = abc sin Z sin w. 

Suppose a sphere to be described having for its centre the vertex of the 
trihedral angle whose edges are a, 6, and c. The spherical triangle whose 
vertices are the points where a, 6, and c meet the surface has for its sides 
l y m, n; and w is the perpendicular arc to the side I from the opposite 
vertex. 

Let X, M, N denote the angles of the triangle opposite I, m, n, 
respectively. 

Then, by [39], 

sin w = sin m sin N 
By [12] , = 2 sin m sin i N cos £ N. 

Let s = i(l + m + n). 

By [47], sin -§- N = V sin (s — I) sin (s — m) esc I esc m, 
and 



cos i JV = Vsin s sin (s — n) esc Z esc m 
.-. sin w — 2 sin m 
2 sin m 



'V 5 



sin (s — Z) sin (s — m) 



sin Z sin m 



Vsin s sin ( 
sin Z si 



(s-n) 



sin Z sin m 

2 



Vsin s sin (s — Z) sin (s — m) sin (s — n) 



Vsin s sin (s — Z) sin (s — ?n) sin (s — n). 
sin Z 

V = abc sin Z sin w 

= 2 a&c Vsin s sin (s — Z) sin (s — m) sin (s — n). 



TEACHERS' EDITION. 



279 



4. The continent of Asia has 
nearly the shape of an equilateral 
triangle, the vertices being the East 
Cape, Cape Romania, and the Prom- 
ontory of Baba. Assuming each 
side of this triangle to be 4800 
geographical miles, and the earth's 
radius to be 3440 geographical miles, 
find the area of the triangle : (i) re- 
garded as a plane triangle ; (ii) re- 
garded as a spherical triangle. 



(i) Area 
Altitude 



= -J- (base x altitude). 
= V4800 2 - 2400 2 



= 2400 V4 - 1 
= 2400 V3. 



log 2400 : 

lOg V3: 
log 2400 : 

log area : 
Area : 



: 3.38021 

0.23856 

: 3.38021 

: 6.99898 

: 9976500. 



(ii) 



a = b = c -. 

2S: 
S : 

i (s — a) - 

i(S-b): 

i(s-c) 

log tan i s : 

log tan i{s — a) - 

log tan i(s — b) - 

log tan i(s — c) ■- 

log tan 2 %JE: 

log tan i E ■- 
E 



ttW. 

80°. 



E 
180' 

: 4800' 

: 240°. 
: 120°. 
: 60°. 

: 20°. 
: 20°. 
: 20°. 

: 10.23856 

9.56107 

: 9.56107 

: 9.56107 

: 8.92177 

: 9.46088 

:16° V 7.5 /y 
: 64° 28 r 30" 
: 232110". 



log 



\ogE = 5.36570 
= 4.68557 

.07312 



648000 
\ogR 2 = > 
\ogF= 7.12439 

F = 13316560. 



5. A ship sails from a harbor in 
latitude I, and keeps on the arc of a 
great circle. Her course (or angle 
between the direction in which she 
sails and the meridian) at starting 
is a. Find where she will cross the 
equator, her course at the equator, 
and the distance she has sailed. 

Let NESWbe the earth, WCE 
the equator, N and S the north and 
south poles. Let A be the point 




from which the ship starts, AFB 
the parallel of latitude of A, and 
AB the great circle of the ship's 
course. 




280 



SPHERICAL TRIGONOMETRY. 



Then 
BAE — a — course of ship, 
AE = I = latitude of its starting 

place, 
BE — m — place of crossing the 

equator, 
90° — B = course at equator, 
AB = d = distance sailed. 



By Napier's Rules 



and 



sin I — tan m cot a, 
cos B = cos Z sin a, 
cos a = tan Z cot d. 



Whence, tan m = sin Z tan a, 

cos B == cos Z sin a, 

and cot d = cot Z cos a. 



6. Two places have the same latitude Z, and the distance between the 
places, measured on an arc of a great circle, is d. How much greater is 
the arc of the parallel of latitude between the places than the arc of the 
great circle ? Compute the results for I = 45°, d = 90°. 




Let B and C be the two given places of latitude Z, d the arc of the 
great circle passing through B and (7, and k the arc BC of the parallel of 
latitude of B and C. 

By Prob. 1, Ex. XXXV, 

sin i A = sin \ d esc (90° - I) 

= sin \ d sec I. 
.-. -J- A = shr -1 (sin \ d sec I). 
A = 2 sin- 1 (sin i d sec I). 
A = arc a. 
. arc k = a cos I 
= A cos Z 

= 2 cos Z sin -1 (sin -J- d sec Z). 
Zc — d = 2 cos I sin- x (sin -J d sec Z) — cZ. 
k - d = 2 cos 45° sin-i (sin 45° sec 45°) - 90° 

= 2xiV2sin-if-^i^_ 9 0° 
V cos 45°/ 

= V / 2sin- 1 (l)-90° 

= V2 x 90° - 90° 

= 90°(V2-i). 



Again, 



If Z = 45° and d = 90°, 



TEACHERS' EDITION. 



281 



7. The distance d between two 
places and the latitudes I and V of 
the places are known. Find the 
difference between their longitudes. 




Let C represent the north pole, 
A the position of one place, B 
the position of the other, and arc 
AB = d. 

If the latitudes of A and B are 
I and T, 

AC = 90° -I, 
BC = 90° - V. 

By [47], taniC = 
Vsec s sec (s — d) sin (s — I) sin (s — V) , 
where 2 s = I + V -f & 

8. Given the latitudes and longi- 
tudes of- three places on the earth's 
surface, and also the radius of the 
earth ; show how to find the area of 
the spherical triangle formed by arcs 
of great circles passing through the 
three places. 

The sides of the triangle are found 
by Sect. LXV ; and the area is found 
from the sides by Sect. LXIII. 

9. The distance between Paris and 
Berlin (the arc of a great circle) is 
equal to 472 geographical miles. The 
latitude of Paris is 48° 50' 13"; that of 
Berlin, 52° 30' 16". When it is noon 
at Paris, what time is it at Berlin ? 



Let A O represent the latitude of 
Paris, and BK the latitude of Berlin. 
Then C represents the difference in 
longitude. 




CA = b = 41° 9' 47" 
CB = a = 37° 29' 44" 
AB=c= 7° 52' (472 - 60) 
2 s = 86° 31' 31" 

s = 43° 15' 45.5". 
s-a= 5° 46' 1.5". 
s-b= 2° 5' 58.5". 
s - c = 35° 23' 45.5". 

By [47], tanHC = 
esc s sin (s — a) sin (s - b) esc (s — c). 

log esc s= 0.16409 

log sin (s- a) = 9.00210 

log sin (s - b) = 8.56391 

log esc (s - c)= 0.23715 

logtanHC = 17.96725 

logtan-£-C= 8.98363. 

iC= 5° 30' 2". 
= 11° 0'4". 
15 )11° 0' 4" 

44 min. y 4 - sec. 

Therefore, the time at Berlin is 
12 hr. 44 min. p.m. 

10. Given tfie altitude of the pole 
45°, and the azimuth of a star on 
the horizon 45° ; find the polar dis- 
tance of the star. 



282 



SPHERICAL TRIGONOMETRY. 



Let Z be the zenith, P the pole, 
and M the position of the star. In 
the spherical triangle ZMP, 




ZP = 90° - I = 45°, 
ZM = z = 90°, 
Z = a = 45°. 

Required p. 

By [45], 

cosp = cos (90° — I) cos z 

+ sin (90° — I) sin z sin a 

= cos 45° cos 90° 

+ sin 45° sin 90° sin 45° 
= sin2 45° 

= i- . 
.-. p = 60°. 

11. Given the latitude I of the 
observer, and the declination d of the 
sun ; find the local time (apparent 
solar time) of sunrise and sunset, 
and also the azimuth of the sun at 
these times (refraction being neg- 
lected). When and where does the 
sun rise on the longest day of the 
year (at which time d = -f 23° 27') 
in Boston (I = 42° 21'), and what is 
the length of the da^ from sunrise 
to sunset ? Also, find when and 
where the sun rises in Boston on 
the shortest day of the year (when 



d = - 23° 27'), and the length of 
this day. 

(i) To find the heur angle t when 
the sun is on the horizon. 




PM=90°-d, 
ZQ = 90°, 
PQ = l 

Then in triangle PMQ, by [40], 
cos QPM = tan PQ cot PM, 
or cos t — — tan I tan d. 



Time of sunrise 
15 



(12 ) o'clock A.M. 

\ 15/ 



Time of sunset 

■(A)- 



clock P.M. 



(ii) To find azimuth a = MQ. 
By [38], 

cos PM = cos PQ cos QJf, 
or sin d = cos I cos a. 

.-. cos a = sin d sec I. 

(iii) At Boston on the longest day 
cos t = — tan d tan Z. 
log tan d = 9.63726 
log tan I = 9.95977 
log cos t = 9.59703 (n) 
t = 113° 17 7 26" 






TEACHERS' EDITION. 



283 



— = 7 hr. 33 min. 10 sec. 
15 

12 = 4 hr. 26 min. 50 sec. 

15 

Length of longest day 

2t 
= — = 15 hr. 6 min. 20 sec. 
15 

cos a = sin d sec I. 

log sin d = 9.59983 
log sec I = 0.13133 
log cos a = 9.73116 



a : 



57° 25' 15" 



(iv) At Boston on the shortest day 

cos t = tan d tan I. 

t = M° 42' 34". 

t 

15 
t 



12 



15 



■ = 4 hr. 26 min. 50 sec. 
7 hr. 33 min. 10 sec. 



; 8 hr. 53 min. 40 sec. 



Length of shortest day 
_2t 
~15 

cos a' ' = — sin d sec I. 
.-. a' = 180° - a 

= 122° 34' 45". 

12. When is the solution of the 
problem in Example 11 impossible, 
and for what places is the solution 
impossible ? 

The solution is impossible if 
cos t > 1 or < — 1 or if cos a > 1, 
or< — 1, i.e., if (for positive decli- 
nation) 

tan I > cot d, 
or sin I > cos d ; 

that is, if I > 90° - d. 



The maximum value of d is 23° 
27' ; hence the minimum value of I 
is 66° 33". The solution is therefore 
impossible only for places within the 
Arctic or Antarctic circles. For such 
places at certain seasons depending 
on d the sun fails to rise during 24 
hours. 

13. Given the latitude of a place 
and the sun's declination ; find his 
altitude and azimuth at 6 o'clock 
a.m. (neglecting refraction). Com- 
pute the results for the longest day 
of the year at Munich (I = 48° 9'). 
In the spherical triangle PZM, 
PZM = a, 
PZ = 90° - I, 
PM = 90° - d, 
ZPM = t = 90°, 
ZM = 90° - h. 
The declination of the sun on the 
longest day is 23° 27'. 




By [38], 

cos MZ — cos PZ cos MP. 
cos (90°- h) 

= cos (90° -I) cos (90° 



-d). 



284 



SPHERICAL TRIGONOMETRY. 



.-. sin h = sin I sin d. 
By [42], 

sin PZ = tan PM cot PZM. 
sin (90° - I) 

= tan (90° - d) cot a. 
cos Z = cot d cot a. 
.-. cot a = cos Hand. 
If Z = 48° 9' and d = 23° 27', 
log sin I =9.87209 
log sin d = 9.59983 
log sin h = 9.47192 
Altitude = h = 17° 14 / 35". 
log cos I =9.82424 
log tan d = 063726 
log cot a = 9.46150 
Azimuth = a = 73° 51' 34". 

14. How does the altitude of the 
sun at 6 a.m. on a given day change 
as we go from the equator to the 
pole ? At what time of the year is 
it a maximum at a given place? 
(Given sin h = sin I sin d.) 

The farther the place from the 
equator, the greater the sun's alti- 
tude at 6 a.m. in summer. At the 
equator it is 0°. At the north pole 
it is equal to the sun's declination. 
At a given place, the sun's altitude 
at 6 a.m. is a maximum on the 
longest day of the year, and then 
sin h = sin I sin e (where e = 23° 27'). 

15. Given the latitude of a place 
north of the equator, and the dec- 
lination of the sun ; find the time of 
day when the sun bears due east 
and due west. Compute the results 
for the longest day at St. Peters- 
burg (I = 59° 56'). 




PM = 90° - d. 
PZ = 90° - I 
PZM = 90°. 
By [40], 
cos ZPM = tan PZ cot PM. 

cost = tan (90° -I) cot (90° -d). 
.-. cos t = cot I tan d. 

The times of bearing due east and 
due west are 

12 a.m. and — p.m., 

15 15 

respectively. 

At St. Petersburg on the longest. 

day I = 59° 56', d = 23° 27'. 

log cot I = 9.76261 
log tan d = 9.63726 
log cos t =9.39987 

t = 75° 27' 24". 

.-.12 = 6 hr. 58 min. 10 sec. a.ai j 

15 

and — = 5 hr. 1 min. 50 sec. p.m. 
15 

16. Apply the general result in 
Example 15 (cos t = cot I tan d) to 
the case when the days and nights 
are equal in length (that is, when 
d = 0°). Why can the sun in sum- 
mer never be due east before 6 a.m., 



TEACHERS' EDITION. 



285 



or due west after 6 p.m. ? How 
does the time of bearing due east 
and due west change with the decli- 
nation of the sun ? Apply the gen- 
eral result to the cases where I < d 
and I = d. What is it at the north 
pole? 

When the days and nights are 
equal, d = 0°, cos t = 0, and t = 90°; 
that is, the sun is due east at 6 a.m. 
and due west at 6 p.m. Since I and 
d must both be less than 90°, cos t 
cannot be negative ; therefore, t 
cannot be greater than 90°. As d 
increases, t decreases ; that is, the 
times of bearing due east and due 
west both approach noon. 

K I = <*, cos t ; = 1, t = 0°, and the 
times both coincide with noon. 

If l<d, then cos£>l, and the 
case is impossible. 

The explanation of these results 
is that, if d = I, the sun is in the 
zenith at noon, and north of the 
prime vertical at every other time. 
And if d > J, the sun is north of the 
prime vertical the entire day. 

If I > d, the diurnal circle of the 
sun and the prime vertical of the 
place meet in two points, which 
separate farther and farther as I 
increases, the distance between them 
approaching 180° - 23° 27' as I 
approaches 90°. At the pole the 
prime vertical is indeterminate ; but 
near the pole t = 90°, and the sun 
is always east at 6 a.m. 

17. Given the sun-s declination 
and his altitude when he bears due 
east ; find the latitude of the ob- 
server. 




ZM = 90° - h. 
PM = 90° - d. 
PZ = 90° - I 

Since the sun M bears due east, 
MZP is a right angle. 
By [38], 
cos PM = cos PZ cos MZ. 
cos(90°-d) = cos(90 o -0cos(90°-/i). 
.-. sin d = sin I sin h. 
sin I = sin d esc h. 

18. At a point O in a horizontal 
plane MN a staff OA is fixed so 
that its angle of inclination A OB 
with the plane is equal to the lati- 
tude of the place, 51° 30' X., and 
the direction OB is due north. 
What angle will OB make with the 
shadow of OA on the plane, at 
1 p.m., when the sun is on the 
equinoctial ? 




286 



SPHERICAL TRIGONOMETRY. 



Given the direction of OB due 
north, AOB = 51° 30' = Z, and plane 
MN horizontal ; find BOC. 

SPZ — hour angle of sun at 1 p.m. 
= 15°. 

SPZ — CAB, being vertical angles. 
.•.(MjB=15°. 

ABC = 90°, since OB is the pro- 
jection of OA on plane MN 

Arc AB = 51° 30', being the meas- 
ure of plane angle A OB. 

Then in right spherical triangle 
ABC, by [42], 

tan BC = tan BAC sin J.B. 

log tan BAC = 9.42805 

log sin AB = 9.89354 

log tan BC = 9.32159 

Arc BC = 11° 50' 35". 

.-. BOC = 11° 50' 35". 

19. What is the direction of a 
wall in latitude 52° 30° N. which 
casts no shadow at 6 a.m. on the 
longest day of the year ? 




The wall must lie in the plane of 
ZM in order that it may cast no 
shadow. 

PZ = 90° - Z, 
PM = 90° - e, 
P = 90°; 
required MZP = a. 



By [42], 

, sin PZ = tan PM cot PZM. 
sin (90° -l)= tan (90° - e) cot a. 
cos I = cot e cot a. 
.*. cot a = cos Z tan e. 
log cos 1 = 9.78445 
log tan e = 9.63726 
log cot a = 9.42171 

a = 75° 12 r 28". 

20. Find the latitude of the place 
at which the sun rises exactly in the 
northeast on the longest day of the 
year. 

When the sun rises in the north- 
east on the longest day of the year, 
a = 45°, d = 23° 27 7 . 

By Prob. 11, cos a = sin d sec I. 

.-. cos I = sin d sec a. 

log sin d = 9.50983 

log sec a = 0.15051 

log cos I = 9.75034 

I = 55° 45 7 6". 

21. Find the latitude of the place 
at which the sun sets at 10 o'clock 
on the longest day. 

ZPM = 10 x 15° 
= 150°. 
ZM = 90°. 
MP = 90° - I. 
By Prob. 15, 

cos t = cot I tan d. 
.-. cot I = cos t cot d. 
t = 150°. 
d = 23° 27'. 
log cos t= 9.93753 
log cot d = 10.36274 
log cot 1 = 10.30027 

I = 63° 23 7 41". 






TEACHERS 7 EDITION. 



287 



22. To what does the general formula for the hour angle, in Sect. LXX. 
reduce when (i) ft = 0°, (ii) I = 0° and d = 0°, (hi) 3 or d = 90° ? 

By Sect. LXX, sin \ t = ± [cos \ (I -f p + ft) sin -J- (Z -+ p — ft) sec I esc p]* 



By [21], 
(i) Uh = 0° 
By [13], 

By [4], 



= ± [£ (sin {Z + p} — sin /i) sec I esc p] 2 . 
sin££ = :£ [-j-sin(Z -fp)sec Z cscp]^. 
cos t = 1 — 2 sin' 2 £ £ 

= 1 — sin (Z + p) sec Z esc p 

sin Z cos p + cos Z sin p 



= 1 



cos I sin p 
sin Z cosp 



cos £ sin p 
= — tan Zcotp. 
(ii) If I = 0° and cZ = 0°, 

p = 90° - tf 
= 90°. 
.-. sin£$= [i(l -sin ft)]* 
cos t = 1 — (1 — sin ft) 
= sin ft. 
.-. < = 90° - ft 
= z. 
(iii) If Z or cZ = 90°, sec Z or cscp = oo, and the formula is useless. When 
d = 90°, the star is at the pole and its hour angle is indeterminate ; and 
when I = 90°, the place of observation is at the terrestrial pole and the 
meridian is indeterminate. 



23. What does the general for- 
mula for the azimuth of a celestial 
body, in Sect. LXXI, become when 
t = 90° = 6 hours ? 

From Sect. LXXI, 

tan m — cot d cos t, (1) 

and 

tan a = sec (I -f in) tan t sin m. (2) 

Multiply (1) by (2), 

tan a tan m 

— sec (Z + m) cot d sin t sin m. 
.-.tana = sec (Z + m) cot d sin t cos m. 

Here £ = 90°; hence 

tan m == 0, 

and m = 0°. 



.-. tan a = sec Z cot cZ, 
or cot a = cos Z tan d. 

24. Show that the formulas of 
Sect. LXXII, if t = 90°, lead to the 
equation sin I = sin ft esc d ; and that 
if d = 0°, they lead to the equation 
cos I = sin ft sec t. 

From Sect. LXXII, 

tan m = cot d cos £, (1) 

and cos n = cos m sin ft esc cZ. (2) 

(i) If t = 90°, then m = 0° and 
n = 90° - Z ; hence 
cos (90° — Z) = cos 0° sin ft esc cZ. 
sin Z = sin ft esc d. : 



288 



SPHERICAL TRIGONOMETRY. 



(ii) If d = 0°, then m = 90°, n=l. 
Divide (2) by (1), 
cos n cot m = cos m sin ft sec d sec £. 
.*. cos n = sin m sin ft sec d sec t 
.-. cos Z = sin ft sec £. 

25. Given the latitude of the 
place of observation 52° 30 / 16", the 
declination of a star 38°, its hour 
angle 28° 17' 15" ; find the altitude 
of the star. 

By Sect. LXXI, 

tan m = cot d cos t, 
and sin ft = sin (I + m) sin d sec ?n. 
Here d = 38°, 

I = 52° 30' 16", 
£ = 28° 17' 15". 
log cot d = 10.10719 
log cos * = 9.94477 
log tan m = 10.05196 

ro = 48° 25' 10". 

log sin (J -f m) = 9.99206 

log sin d =9.78934 

log sec m = 0.17804 

log sin ft =9.95944 

ft = 65° 37' 20". 

26. Given the latitude of the place 
of observation 51° 19' 20", the polar 
distance of a star 67° 59' 5", its hour 
angle 15° 8' 12" ; find the altitude 
and the azimuth of the star. 

By Sect. LXXI, 
tan m = cot d cos £, 
sin h = sin (I -f m) sin d sec m, 
tan a = sec (I -f- m) tan t sin m. 
Z = 51° 19' 20". 
d = 90° - 67° 59' 5" 

= 22° 0'55". 
* = 15° 8' 12". 



log COt d : 
log COS t : 

log tan m : 

m : 

log sin (I + m) : 

log sin d : 

log sec m -. 

log sin ft : 

ft : 

log sec (I + m) : 

log tan £ : 

log sin m : 

log tan a ■ 

a : 



: 10.39326 
: 9.98466 
: 10.37792 
: 67° 16' 22". 
: 9.94351 
: 9.57386 
0.41302 
: 9.93039 
: 58° 25' 8". 

: 0.32001 (n) 
: 9.43218 
9.96490 
: 9.71709 (n) 
: 152° 28'. 



27. Given the declination of a 
star 7° 54', its altitude 22° 45' 12", 
its azimuth 129° 45' 37 ; find the 
hour angle of the star and the lati- 
tude of the observer. 

In the spherical triangle ZPM, 
sin ZM sin PM 



sin ZPM sin PZM 

.-. sin ZPM 

= sin PZM sin ZM esc PM. 
sin t = sin a sin (90° - ft) esc (90° - d). 
.-. sin t 

— sin a cos ft sec d. 

Here a = 129° 45' 37", 

h= 22° 45' 12", 
d = 7° 54'. 
log sin a = 9.88577 
log cos ft = 9.96482 
log seed = 0.00414 . 
log sin t =9.85473 
t = 45° 42'. 
By Sect. LXXII, 
tan m = cot d cos t, 
cos n = cos m sin ft esc d, 
I =90° -(m±n). 



TEACHERS' EDITION. 



289 



log cot (1 : 

log COS t : 

log tan m ■ 

111 : 

log cos m : 

log sin h - 

log esc d ■• 

log cos n : 

n : 

m — n ■■ 

90° -(m-n): 

.\ I: 



10.85773 
9.84411 

10.70184 
: 78° 45' 45" 

9.28976 

9.58745 
: 0.86187 
: 9.73908 
; 56° 44' 39" 
:22° V 6" 
: 67° 58' 54" 
: 67° 58' 54" 



28. Given e = 23° 27' and the 
longitude v of the sun ; find the 
declination d and the right ascen- 
sion r. 

In the figure let P represent the 
pole of the equinoctial A VB, S the 
position of the sun, and Q the pole 
of the ecliptic EVF. 




Then VS = v, 

VB = r, 

SR = d, 

RVS = e. 

Then in the right triangle RVS, 
by [39], 

sin SB = sinVS x sin BVS, 
or sin d = sin v sin e. 



Also, by [40], 

cos BVS = tan BV cot VS, 
or cos e = tan r cot v. 

.-. tan r = tan v cos e. 

29. Given e = 23° 27', the lati- 
tude of a star 51°, its longitude 
315°; find its declination and its 
right ascension. 




Here VT = 315°, 
TM = 51°, 
BVT= 23° 27'; 
to find VB = r, 
and BM = d. 

In the right triangle VTM, 
By [38], 

cos VM = cos VT cos TM, 
By [42], 

tan MVT = tan Jf T esc VT. 

log cos FT= 9.84949 
log cos T3f= 9/79887 
log cos IOf= 9.64836 

VM = 63° 34' 36". 
log tan TM= 10.09163 
log esc VT = 0.15051 (n) 
log tan MVT = 10.24214 (n) 



290 



SPHERICAL TRIGONOMETRY. 



i T/FT = -(60°12 / 14.5 // ). 

In the right triangle RVM, 

RVM=RVT + TVM 

= 23° 27' -(60° 12' 14.6") 
= -(36° 45' 14.5"). 

By [39], 

sin RM = sin VM sin RVM. 

log sin Of = 9.95208 

log sin R VM = 9.77698 

log sin RM= 9.72906 

RM = d = 32° 24' 12". 

Also, by [42], 

sin VR = tan Elf cot RVM. 

log tan RM =9.80257 
log cot RVM = 0.12677 (w) 
log sin VR = 9.92934 (w) 

FB='-(68°11'43"). 

.-. VR = 360° - 58° 11' 43" 
= 301° 48' 17". 



30. Given the latitude of the ob- 
server 44° 50', the azimuth of a star 
138° 58', its hour angle 20° ; find its 
declination. 




Given c = 90° - 44° 50' 
= 45° 10', 
A = 138 6 58', 
B = 20°. 
i(A-B) = 59° 29'. 
i (A + B) = 79° 29'. 
ic = 22°35 / . 
log cos i (A -B) = 9. 70568 
log sec £(.4 + B) = 0.73869 
log tan | c = 9.61901 
log tan i(a + b) = 0.06338 

f(a + &) = 49°9 / 57.7" 

log sin \ (A - B) = 9.93525 

log esc $ (A + B) = 0.00736 

log tan \ c = 9.61901 

log tan £ (a- b) = 9.56162 

i(a-6) = 20°l'24.6" 
... a = 69° ir 22". 
d = 90° - 69° ir 22" 

= 20° 48' 38". 



31. Given the latitude of the place of observation 51° 31' 48", the alti- 
tude of the sun west of the meridian 35° 14' 27", its declination 4- 21° 27' ; 
find the local apparent time. 

By Sect. LXX, 

sin i t = Vcos i (I + p + h) sin £ (I + P — h) esc p sec I. 
Here I = 51° 31' 48", 

h = 35° 14' 27", 
p = 90° - 21° 27' = 68° 33'. 






teachers' edition. 



291 



*(l +p + *) = 77° S^ 87.fi". 

i(l+p -ft) = 42° 25' 10.5". 

log cos | (I + p + ft) = 9.32982 

log sin i (I + p - ft) = 9. 82901 

log csc p = 0.03117 

log sec I = 0.20614 

logsinHt= 9.39614 

log sin -H= 9.69807 

££ = 29° 55' 54.5". 
£ = 59° 51' 49". 

— = 3 hr. 59 min. 27 T 4 T sec. p.m. 
15 l0 

32. Given the latitude of place /, the polar distance p of a star, and 
its altitude ft ; find its azimuth a. 




Altitude = ZJf= 90° -ft. 
Co-latitude = PZ = 90° - I 
Polar distance == PJf = 90° 
Azimuth = PZM = a. 
cosiA 



d = p. 



By [47], cos i A = Vsin s sin (s — a) esc b esc c. 

Let A = PZM or a, 
« = p, 
6 = 90° - A, 
c = 90° - I. 
Then sin s = sin [90° -i(l + h-p)] 

= cos £(Z + ft — p). 
sin (s - a) = sin [90° - £ (Z + ft + p)] 
= cos£(Z + ft + p). 
csc 6 = esc (90° — ft) = sec ft. 
esc c = csc (90° — I) = sec Z. 
.-. cos i a = Vcos £ (I -f ft + p) cos J- (I + ft — p) sec Z sec ft. 



SURVEYING. 



Exercise II. Page 230. 



2. From the bearings obtained in 
Example 1 find the value of each of 
the interior angles. What is their 
sum ? 

Since the given field has 5 sides 
the sum of the interior angles is 
(Geometry, §205) 

3 x 180° = 540°. 

4. Kange out a line whose bear- 
ing is N. 38° 30' W., and at some 



point in this line range out another 
line making a right angle with it. 
What is the bearing of the second 
line? 

The second line makes an angle 
of 90° with the first. 

90° - 38° 30' = 51° 30'. 

Therefore, the bearing of the sec- 
ond line is N. 51° 30' E. 



Exercise III. Page 235. 



2. Lay out the entire angular 
magnitude about some point into 
four or more angles, and measure 
each of them. What should the 
sum of them equal ? 

The sum of them should equal 
360°. 



3. If the constant of a transit, 
adjusted to one foot 100 feet away, 
is 3.8 inches, what is the true length 
of a line when the indication on the 
rod is 2.35 feet ? 

The true distance is 235 feet 3.8 
inches. 



Exercise V. Page 263. 

1. Required the area of a triangular field whose sides are 13 chains, 
14 chains, and 15 chains. 



Area = Vs(s — a) (s — b) (s — c) 
= V21 x 8 x 7 x 6 sq. ch. 
= V3 2 x 7 2 x 2 4 sq. ch. 
= 84 sq. ch. 
= 8.4 a. 
= 8 a. 64 p. 
292 



teachers' edition. 293 

2. Required the area of a triangular field if it has two angles 48° 30' 
and 71° 45', and the included side 20 chains. 

_ a 2 sin B sin C 
~2sin(J5+ C) 

= — smBsinCcsc(B+ C) 

2 

= 200 sin 48° 30' sin 71° 45' esc 120° 15'. 

log 200 = 2.30103 

log sin 48° 30' = 9.87446 

log sin 71° 45' = 9.97759 

log esc 120° 15' = 0.06357 

log F= 2.21665 

F= 164.68. 

Area = 164.68 sq. ch. = 16.468 a. = 16 a. 74|f p. 



3. Required the area of a triangular field whose base is 12.60 chains, 
and altitude 6.40 chains. 

Area = £(12.6 x 6.4) sq. ch. 
= 40.32 sq. ch. 
= 4.032 a. 
= 4 a. 5 T 3 7 P. 

4. Required the area of a triangular field which has two sides 4.50 
chains and 3.70 chains, and the included angle 60°. 

Area = \ be sin A 

= i x 4.5 x 3.7 x 0.866 = 7.20945. 
Area = 7.20945 sq. ch. 

= 0.720945 a. 

= 115 ^ p. 



5. Required the area of a field in the form of a trapezium, one of 
whose diagonals is 9 chains, and the two perpendiculars upon this diag- 
onal from the opposite vertices 4.50 chains and 3.25 chains. 

Area = 9 x |(4.5 + 3.25) sq. ch. 
= 34.875 sq. ch. 
= 3.4875 a. 
= 3 a. 78 p. 



294 



SURVEYING. 



6. Required the area of the field ABCBEF (Fig. 36), if AE = 9.25 
chains, FF' = 6.40 chains, BE = 13.75 chains, DD' = 7 chains, BB - 10 
chains, CC = 4 chains, and AA' = 4.75 chains. 




2 area J.F.E : 
2 area #DjE; JL 
2 area jBDC 



6.4 x 9.25 = 59.2 

: 13.75(4.75 + 7) = 161.5625 
: 10 x 4 = 40. 



2 area ABCDEF 
area ABCBEF 
130.38125 sq. ch. = 13.038125 a. 



= 260.7625 
= 130.38125 
= 13 a. 6 T \?< 



7. Determine the area of the field ABCB from two interior stations, 
P and P', if PP' = 1.50 chains, 




PP'C = 89° 35', 
PP'B = 185° 30', 
PPOl = 309° 15', 
PP f B = 349° 45', 



P'PB = 3° 35', 
P'PA = 113° 45', 
P'PB = 165° 40', 
P'PC = 303° 15'. 



Area = A P.4D + A PCB + A PBC + A P^i£. 



ZPP'D= 10° 15', 
Z PDP' = 4° 5', 
Z PP'JS = 174° 30', 



Z PP'^i = 50° 45', 
ZP^4P' = 15° 30', 
Z PBP' = 1° 55', 



Z P'PC = 56° 45', 
Z PCP' = 33° 40'. 



TEACHERS EDITION. 



295 



PD 



PP' sili PP'D 



sin PDP' 

log PP' =0.17609 

log sin PP'D = 9.25028 

colog sin PDP' = 1.14748 

log PD =0.57385 



PA 



PP' sin PP'A 



sin PAP' 

log PP' =0.17609 

log sin PP'A = 9.88896 

colog sin PA P' = 0.57310 

loo- PA =0.63815 



PC = 



PP' sin PP'C 



sin PGP' 

log PP' =0.17609 

log sin PP'C = 9.99999 

colog sin PCP' = 0.25621 

log PC =0.43229 



PB 



PP' sin PP'B 



sin PBP' 

logPP' =0.17609 

log sin PP'D = 8.98157 

colog sin PBP' = 1.47566 

logPD =0.63332 



Z APB = 51° 55', Z DPC = 137° 35', Z BPC = 60° 20', Z APB = 110° 10' 



2 area P^ID = PD x PA sin APD. 
logPD =0.57385 

log PA =0.63815 

log sin A PD = 9.89604 
log 2 area PAD = 1. 10804 
2areaP^iD =12.824. 



2 area PCD = PD x PC sin DPC. 
logPD =0.57385 

log PC = 0.43229 

log sin DPC = 9.82899 
log 2 area PCD = 0.83513 
2 area PCD =6.8412. 



2 area PBC = PC x PB sin DPC. 
log PC =0.43229 

logPD =0.63332 

log sin BPC = 9.93898 
log 2 area PBC =1.00459 
2 area PBC =10.106. 



2 area P^iD = PA x PB sin APB. 
logP^l =0.63815 

logPD =0.63332 

log sin APB = 9.97252 
log 2 area P^ID = 1.24399 
2 area P.4D =17.538. 



2AP^1D=12.824 
2 A PCD = 6.8412 
2 A PBC =10.106 
2APAB = 17.538 
2ABCD =47.3092 

ABCD =23.6546. 
23.6546 sq. ch. = 2.36546 a. = 2 a. 58£p,, nearly. 



296 



SURVEYING. 



8. Required the area of the field ABCDEF (Fig. 37), if AF' = 4 chains, 
FF' = 6 chains, EE' = 6.50 chains, AE' = 9 chains, AD = 14 chains, AC 
= 10 chains, AB' = 6.50 chains, BB' = 7 chains, CC = 6.75 chains. 




2 area AFF' =4x6 =24. 

2 area F'E'EF = 5 (6 4- 6.5) =" 62.5 

2 area EE'D =6.5x5 = 32.5 

2 area ^£5' =6.5x7 = 45.5 
2 area BCCB' =3.5(7 + 6.75) = 48.125 

2 area CDC 7 = 6.75 x 4 = 27. 

2 area ABCDEF = 239.625 

area, ABCDEF = 119.8125 

119.8125 sq. ch. = 11.98125 a. = 11 a. 157 p. 




9. Required the area of the field 
AGBCD (Fig. 32, p. 260), if the 
diagonal AC = 5, BB' (the per- 
pendicular from B to AC) = 1, 
DD' (the perpendicular from D 
to AC) = 1.60, ^J£' = 0.25, FF' 
= 0.25, ££' = 0.60, HH' = 0.52, 

inr = 0.54, ..or = 0.2, E'F' = 

0.50, F^ = 0.45, <?'#' = 0.46, 
S'lT = 0.60, and 2TJ5 = 0.40. 

= 5(1 + 1.6) =13. 

= 0.25 x 0.2 = 0.05 

= 0.5(0.25 + 0.25)= 0.25 
= 0.45(0.25 + 0.6)= 0.3825 

0.504 

0.636 



2 area ADCB 

2 area AEE' 

2 area EE'F'F 

2 area FF'G'G 

2 area GG'H'H = 0.45 (0.6 + 0.52) = 

2 area HH'K'K =0.6 (0.52 + 0.54) 



2 area KK'B = 0.4 x 0.54 



= 0.216 



2 area ADCBKHGFE 
area ADCBKHGFE 



15.0385 
: 7.51925. 






TEACHERS EDITION. 



297 



10. Required the area of the field AGBCD (Fig. 33, p. 260), if AD = 3, 
AC = 5, AB = 6, angle I) AC = 45°, angle BAC = 30°, AW - 0.75, AF' 
= 2.25, AH = 2.53, AG' = 3.15, EE' = 0.60, PF' = 0.40, and GG' = 0.75. 




2 area ^4DC f = 3 x 5 x sin 45° = 3 x 5 x 0.7071 = 10.6065 
2 area ABC = 5 x 6 x sin 30° =5x6x0.5 =15. 
Z&ve&HGB = 0.75 x (6 - 2.53) = 0.75 x 3.47 = 2.6025 



. 2nrenADCBGHA 
2 area ^L J£E" =0.75x0.6 
2 area EE'F'F = 1.5 (0.6 + 0.4) 
2 area PF'# = 0.4 x 0.28 

. 2 area AEFHA 

. 2 area, AGBCD 
area -46? P CD 



0.45 

1.5 

0.112 



= 28.209 



2.062 

26.147 

: 13.0735 



11. Determine the area of the field ABCD from two exterior stations 
P and P', if PP' = 1.50 chains, 

p' P 




P'PP = 41° 10', 
P'P^l = 55° 45', 
P'PC = 77°20 / , 
P'PD = 104° 45', 
Area = (A P'CB + A P'CD) 



PP'D = 66° 45', 
PP / C= 95° 40', 
PP'B = 132° 15', 
PP'^4 = 103° 0'. 

(AFi^ + AP'iD). 



298 



SURVEYING. 



Z PBP' = 6° 35', 
Z PDP' = 8° 30', 



ZPCP'= 7° 0', 
Z PAP' = 21° 15'. 



P'B = 



PP' sin P'PB 



sin PBP' 

logPP / = 0.17609 

log sin P'P£ = 9.81839 

colog sin PBP' = 0.94063 

logP'£ = 0.93511 

PP^sinP^PC 

sin PCP' 

\ogPP' =0.17609 

log sin P'PC = 9.98930 

colog sin PCP' = 0.91411 

log P'C =1.07950 

Z BP'C = 36° 35', 
Z CP'D = 28° 55', 

2 area P'CB = P'C X P'£sin ^P'C. 

log P'C =1.07950 

logP'.B =0.93511 

log sin BP'C = 9.77524 
log 2 area P'CB = 1.78985 
2 area P'CB = 61.639. 

2 area P'A£ = P'J3 x P'^i sin^iP'£. 
logP'£ =0.93511 

logPOi =0.53415 

log sin AP'B = 9.68897 
log 2 area P0i£ = 1. 15823 
2areaP^i£ = 14.396. 



P'B: 



PP' sin P'PD 



sin PDP' 
logPP' = 0.17609 
log sin P'PD = 9.98545 
colog sin PBP' = 0.83030 
log P'D = 0.99184- 



P'A = 



PP' sin P'PA 



sin P^IP' 

logPP' =0.17609 

logsinP'P^l = 9.91729 

colog sin P.4P' = 0.44077 

logP'J. =0.53415 

Z AP'B = 29° 15', 
Z AP'B = 36° 15'. 

2 area P'CD = P'C x P'Dsin CP'D. 
log P'C =1.07950 

logP'D =0.99184 

log sin CP'D = 9.68443 
log 2 area P' CD = 1.75577 
2 area P' CD =56.986. 

2 area PMD = P'^l x P'Dsin AP'D. 
logP'A =0.53415 

logP'D =0.99184 

log sin AP'D = 9.77181 
log 2 area P'AD = 1.29780 
2 area P'.4D = 19.852. 



. 2 area ABCD = 61.639 sq. ch. + 56.986 sq. ch. - (14.396 + 19.852) sq. ch. 

= 118.625 sq. ch. - 34.248 sq. ch. 

= 84.377 sq. ch. 
.-. area ABCD = 42.1885 sq. ch. 

= 4.21885 a. 

= 4 a. 35 p., nearly. 



TEACHERS' EDITION. 



299 



12. Find the area of the field ABODE (Fig. 35, p. 202) if the co-ordi- 
nates, in chains, of the vertices taken in order are (1.40, 6.75), (4.60, 8.32), 
(9.00, 9.05), (12.15, 5.58), and (5.27, 1.16). 

C 



Y 


a(^ 


B /^ 




I \ 

i 

1 
1 
i 
I 

i ^^. 

i ^^^^ 


2^ D 




r' 






E\ 

i 


i 

i 




Y- 


X O 


L 


M 


N 


P 


R 


-A 


Y 















6.75)} 



ABODE = i {xi (y 6 - y 2 ) + x 2 (y t - y 3 ) + x 3 (y 2 - Vi) 
+ X4(y B -y&) + z B (y4--y 1 )} 

= 1(1.40(1.16 - 8.32) + 4.60(6.75 - 9.05) 

+ 9 (8.32 - 5.58) + 12.15 (9.05 - 1.16) + 5.27 (5.58 
= i(- 10.0240 - 10.5800 + 24.6600 + 95.8635 - 6.1659) 
= i(120.5235 -26.7699) 
= }X 93.7536 
= 46.8768. 
46.8768 sq. ch. = 4.68768 a. = 4 a. 110 r. 

13. Find the area of the field ABODE (Fig. 35, p. 262) by measuring 
distances as follows : AL = 400 feet ; BM = 700 feet ; CP = 680 feet ; 
DR = 380 feet ; EN = 200 feet ; LM = 150 feet ; MN = 250 feet ; NP 
= 200 feet ; PR = 220 feet. 

2 area ABODE = 2 A BML + 2 BCPM + 2 ODRP - 2 AENL - 2 EDRN. 
2 area^JiX = LM (AL + BM) = 150(400 + 700) = 165.000. 
2 area BCPM = MP (BM + CP) = 450 (700 + 680) = 621,000. 
2 area CDRP = PR (CP + DR) = 220 (680 + 380) = 233,200. 
2 area AENL = LN (AL + EN) = 400 (400 + 200) = 240,000. 
2 area EDRN = NR (EN + DR) = 420 (200 + 380) = 243,600. 
.-. 2 2irz%, ABODE =(165,000 + 621,000 + 233,200 -240,000- 243,600) sq. ft. 
= (1,019,200 - 483,600) sq. ft. 
= 535,600 sq. ft, 
.-. area ABODE = 267,800 sq. ft. 

— _2 6_7_8_0_0. A 

— ¥3560 A ' 

= 6.148 a. 
= 6a.231|p. 



300 








SURVEYING. 














Exercise VI. Page 272. 










1. 






o 

H 
< 
H 


Bearing. 


as 

s 


N. 


S. 


E. 


W. 


M.D. 




N.A. 


S.A. 


1 

2 


S. 75° E. 
S. 15° E. 


6.00 
4.00 




1.55 
3.86 


5.79 




5.79 
6.83 


5.79 
12.62 




8.9745 
48.7132 


1.04 


3 
4 


S. 75° W. 

N.45QE. . 


6.93 
5.00 


3.54 


1.80 


3.54 


6.70 


0.13 
3.67 


6.96 
3.80 


13.4520 


12.5280 


' "1.T9 




5 


N. 45o W. 


5.191 


3.67 




10.37 


3.67 


0.00 


3.67 


13.4689 






7.21 


7.21 


10.37 




2G.9209 


70.2157 
26.9209 




21.647 sq 


ch. = 2.1647 A. = 2 A. 26 P., nearly. 




43.2948 










21.6474 

























O 

H 

H 

H 
02 


Bearing. 


s 


N. 


S. 


E. 


W. 


M.D. 


ft 
A 


N.A. 


S.A. 


1 


N.450E. 


10.00 


7.07 




7.07 




7.07 


7.07 


49.9849 




2 


S. 75o E. 


11.55 




2.99 


11.16 




18.23 


25.30 




75.6470 


3 


s. 150 W. 


18.21 




17.59 




4.71 


13.52 


31.75 




558.4825 


4 


N. 45° W. 


19.11 


13.51 






13.52 


0.00 


13.52 


182.6552 








20.58 


20.58 


18.23 


18.23 




232.6401 


634.1295 
232.6401 




200.74 sq. 


ch. = 20.074 A. = 20 A. 12 P., nearly. 




401.4894 










200.7447 



teachers' edition. 



301 



Bearing. 


s 


N. 


S. 


E. 


W. 


N. 51° 45' W. 
S. 85° W. 

S. 55° W W. 


2.39 
6.47 
1.62 


1.48 


0.56 
0.93 




1.88 
6.45 
1.33 




1.49 
1.48 




9.66 








0.01 







O 

< 


Bearing. 


EB 

s 


N. 


S. 


E. 


W. 


M.D. 




N.A. 


S.A. 


2 
3 
4 
5 
6 
1 


S. W. 
N. 3°45'E. 
S. 66° 45' E. 
N.15° E. 
S. 82° 45' E. 
S. 2° 15' E. 


6.39 
1.70 
4.98 
6 03 
9.68 


6.36 
4.80 


0.03 


0.43 


9.65 


9.65 
9.22 
7.66 
6.37 
0.39 
0.00 


9.65 

18.87 

16.88 

14.03 

6.76 

0.39 


120.0132 
67.3440 


0.2895 

11.3096 

5.2052 
3.7791 


0.67 
0.77 




1.56 

1.29 

5.98 
0.39 




9.69 






11.16 


11.16 


9.65 


9.65 




187.3572 
20.5834 


20 5834 




83.39 sq. 






cb.= 


5.339 A 


.= 8 A 


54 P., 


nearly 






166.7738 
83.3869 



302 



SURVEYING. 



ft 

o 

H 


Bearing. 


H 
S 


N. 


S. 


E. 


W. 


M.D. 


A 
A 


N.A. 


S.A. 


3 
4 

1 

2 


S. 3°00'E. 
E. 

N. 5°30'W. 
S. 82° 30' W. 


5.33 
6.72 
6.08 
6.51 


0.03 


5.29 


0.28 
6.73 


0.57 


0.28 
7.01 
6.44 
0.00 


0.28 

7.29 

13.45 

6.44 


0.2187 
81.7760 


1.4812 

5.2808 


0.82 


6.08 




6.44 








6.11 


6.11 


7.01 


7.01 




81.9947 
6.7620 


6.7620 




37.61 sq. 


ch.= 3 


.761 A. 


= 3A. 


122 P., 






nearly 




75.2327 
37.6163 



5. 



© 

H 

H 
02 


Bearing. 


s 


N. 


S. 


E. 


W. 


M.D. 


% 
A 


N.A. 


S.A. 


3 
4 
1 
2 


S. SOOO'E. 
N.88°30'E. 
N. 6°15'W. 
S. 81° 50' W. 


5.86 
4.12 
6.31 
4.06 


0.12 


5.83 


0.53 


0.67 


0.53 
4.67 
4.00 
0.00 


0.53 
5.20 

8.67 
4.00 


0.6240 
54.4476 


3.0899 

2.2800 


0.57 

~Ct5S- 


4.14 


6.28 






4.00 
-4£§- 




6.40 


6.40 


4.67 


4.67 




55.0716 
5.3699 


5.3699 




24.85 sq 


. ch.= 


2.485 A 


.= 2A 






78 P., 


nearly 




49.7017 

24.8508 



TEACHEKS' EDITION. 



303 



6. 



o 




H 












p 






< 
H 

m 


Bearing. 


DQ 


N. 


S. 


E. 


W. 


M.D. 


% 


N.A. 


S.A. 


1 


N. 20° (XK E. 


4.62£ 


4.35 




1.58 




1.58 


1.58 


6.8730 




2 


N.73°00 / E. 


4.16* 


1.22 




3.98 




5.56 


7.14 


8.7108 




3 


S. 45° 15' E. 


6.18£ 




4.35 


4.39 




9.95 


15.51 




67.4685 


4 


S. 38° 3(K W. 


8.00 




6.26 




4.98 


4.97 


14.92 




93.3992 


5 


Wanting. 




5.04 






4.97 


0.00 


4.97 


25.0488 






10.61 


10.61 


9.95 


9.95 




40.6326 


160.8677 
40.6326 




60.12 sq 


ch. = 6.012 A.= 6 A. 2 P., nearly. 


120.2351 










60.1175 



7. 



Bearing. 


DlST. 


N. 


S. 


E. 


W. 


S. 81° 20' W. 
N. 76° 30' W. 


4.28 
2.67 


0.62 


0.65 




4.23 
2.60 




0.65 
0.62 




6.83 








0.03 







Bearing. 


DlST. 


N. 


S. 


E. 


W. 


S. 7° E. 


1.79 




1.78 


0.22 




S. 27o E. 


1.94 




1.73 


0.88 




S. 10O3CKE. 


5.35 




5.26 


0.97 




N. 76° 45' W. 


1.70 


0.39 






1.65 








8.77 


2.07 










39 


1.65 




8.38 


0.42 





304 



SURVEYING. 



to 
o 

H 

< 
H 


Bearing. 


H 
3 


N. 


S. 


E. 


W. 


M.D. 


% 
ft 


N.A. 


S.A. 


1 

2 
3 
4 


S. W. 
N. 5o E. 
S. 870 30/ E. 
S. E. 


8.68 
5.54 


8.65 


0.03 

0.24 

8.38 


0.79 
0.76 
5.56 
5.54 
0.45 
0.42 


6.80 
6.83 


6.80 
6.01 
0.45 
0.00 


680 

12.81 

6.46 

0.45 


110.8065 


0.2040 

1.5504 
3.7730 




8.65 


8.65 


6.80 


6.80 




110.8065 
5.5254 


5.5254 






52.64 sq. 


ch.= . 


5.264 A 


. = 5A 


42 P., 


nearly 






105.2811 
52.6405 



8. 



to" 


H 
< 
H 
02 


Bearing. 


s 


N. 


S. 


E. 


W. 


M.D. 




N.A. 


S.A. 


1 

2 
3 
4 
5 
6 


N. 89° 45' E. 
S. 7°00'W. 
S. 28o 00' E. 
S. 0o45'E. 
N. 84o 45' W. 
N. 2o30'W. 


4.94 
2.30 
1.52 
2.57 
5.11 
5.79 


0.00 
0.02 

0.45 
0.47 
5.76 

5.78 


2.29 

2.28 

1.34 

2.58 
2.57 


4.93 
4.94 

0.71 

0.02 
0.03 


0.29 
0.28 

5.10 
5.09 
0.27 
0.25 


4.93 
4.64 
5.35 
5.37 
0.27 
0.00 


4.93 
9.57 
9.99 
10.72 
5.64 
0.27 


2.5380 
1.5552 


21.9153 
13.3866 
27.6570 




6.21 


6.21 


5.66 


5.66 




4.0932 


62.9595 
4.0932 






29.43 sq. 


ch. = 2 


.943 A. 


= 2 A. 


151 P., 


nearh 


J. 






58.8663 
29.4332 



TEACHERS' EDITION. 



305 



9. An Ohio farm is bounded and described as follows : Beginning at 
the southwest corner of lot No. 13, thence N. 1£° E. 132 rods and 23 liDks 
to a stake in the west boundary line of said lot ; thence S. 89° E. 32 rods 
and 15 T 4 o links to a stake; thence N. 1£° E. 29 rods and 15 links to a 
stake in the north boundary line of said lot ; thence S. 89° E. 61 rods and 
18 t 6 q links to a stake ; thence S. 32^° W. 54 rods to a stake ; thence S. 
35i° E. 22 rods and 4 links to a stake; thence ■ S. 48° E. 33 rods and 
2 links to a stake ; thence S. 7£° W. 76 rods and 20 links to a stake in 
the south boundary line of said lot ; thence N. 89° TV. 96 rods and 10 
links to the place of beginning; containing 85.87 acres more or less. 

Verify the area given and plot the farm. 



H 
Xil 


Bearing. 


s 


N. 


S. 


E. 


TV. 


M.D. 




N.A. 


S.A. 


1 

2 
3 
4 
5 
6 
7 
8 
9 


N. 1°15'E. 
S. 89o E. 
N. 10 15'E. 
S. 89o E. 
S. 32° 30' W. 
S. 35° 15' E. 
S. 48° E. 
S. 7° 30' TV. 
N. 89o TV. 


33.23 

8.154 

7.40 

15.436 

13.50 

5.54 

8.27 

19.20 

24.10 


33.1£ 
33. 2S 

7.3£ 
7.4C 

0.3< 
0.4$ 


5 

> 

> 

) 
> 


0.15 
0.14 

0.29 
0.27 
11.40 
11.39 
4.53 
4.52 
5.54 
5.53 
19.05 
19.03 


0.73 
0.72 

8.15 

0.16 

15.44 
15.43 

3.20 
6.15 


7.24 
7.25 

2.50 

2.51 

24.09 

24.10 


0.73 

8.88 

9.04 

24.48 

17.24 

20.44 

26.59 

24.09 

0.00 


0.73 
9.61 
17.92 
33.52 
41.72 
37.68 
47.03 
50.68 
24.09 


24.2214 

132.4288 

9.3951 


1.4415 

9.7208 
475.6080 
170.6904 
260.5462 
965.4540 




40.96 


40.96 


33 83 


33.83 




166.0453 


1883.4609 
166.0453 








858.7 


sq. < 


:h 


.= 85.$ 


M A. 










1717.4156 

858.7078 



Exercise VII. Page 274. 



1. Find the area of the field ABCD, in which the angle A = 120°, 
B = 60°, C = 150°, and D = 30° ; and the side AB = 4 chains, BC = 
4 chains, CD = 6.928 chains, and DA = 8 chains. 



306 



SURVEYING. 



p 

55 


Bearing. 


CO 

s 


N. 


s. 


E. 


W. 


M.D. 




N.A. 


S.A. 


^b 


N. 


4.000 


4.000 








0.00 


0.00 






^c 


S. 60° E. 


4.000 




2.000 


3.464 




3.464 


3.464 




6.928 


CZ) 


S. 30° E. 


6.928 




6.000 


3.464 




6.928 


10.392 




62.352 


zu 


N.60°W. 


8.000 


4.000 






6.928 


0.00 


6.928 


27.712 






8.000 


8.000 


6.928 


6.928 




27.712 


69.280 
27.712 




20.784 sq. ch. = 2.0784 A. = 2 A. 12£ P., nearly. 


41.568 










20.784 



2. Find the area of the farm ABODE, in which the angle A — 
106° 19', B = 99° 40', = 120° 20', D = 86° 8', and E = 127° 33' ; and 
the side AB = 79.86 rods, BC = 121.13 rods, CD = 90 rods, BE = 
100.65 rods, and EA = 100 rods. 



g 


Bearing. 


M 


N. 


s. 


E. 


W. 


M.D. 


A 


N.A. 


S.A. 


AB 


N. 


79.86 


79.86 








0.00 


0.00 






BC 


N. 80° 20' E. 


121.13 


20.34 




119.41 




119.41 


119.41 


2428.80 




CD 


S. 40° 00' E. 


90.00 




68.94 


57.85 




177.26 


296.67 




20452.43 


DE 


S. 53° 52' W. 


100.65 




59.35 




81.29 


95.97 


273.23 




16216.20 


EA 


N. 73° 41' W. 


100.00 


28.09 






95.97 


0.00 


95.97 


2695.80 






128.09 


128.09 


177.26 


177.26 




5124.60 


36668.63 




1577 


2.02 P. = 98 A. 92 p., nearly. 




5124.60 


31544.03 










15772.02 



teachers' edition. 307 



Exercise VIII. Page 277. 

1. From the square ABCD, containing 6 acres 1 rood 24 perches, part 
off 3 acres by a line EF parallel to AB. 

6 a. 1 r. 24 p. = 64 sq. ch. ; Voi ch. = 8 ch. = AB. 

3 a. =30 sq. ch. 

. n ABFE 30 sq. ch. _ HK _ 

AE = = = 3. i o ch. 

AB 8 ch. 

2. From the rectangle ABCD, containing 8 acres 1 rood 24 perches, 
part off 2 acres 1 rood 32 perches by a line EF parallel to AD which is 
equal to 7 chains. Then, from the remainder of the rectangle part off 2 
acres 3 roods 25 perches by a line GR parallel to EB. 

8 a. 1 r. 24 p. = 84 sq. ch. = ABCD. 
2 a. 1 r. 32 p. = 24.5 sq. ch. = AEFD. 
2 a. 3 r. 25 p. = 29.0625 sq. ch. = EBHG. 

. _ AEFD 24.5 sq. ch. oc _ 

AE = = = 3.o ch. 

AD 7 ch. 

Ar> ABCD 84 sq. ch. ta _ 

AB — = - = 12 ch. 

AD 7 ch. 

EB = AB -AE = 12 ch. - 3.5 ch. = 8.5 ch. 

_ „ EBHG 29.0625 sq. ch. ' . 

EG = = = 3.42 ch., nearly. 

EB 8.5 ch. J 

3. Part off 6 acres 3 roods 12 perches from a rectangle ABCD, con- 
taining 15 acres, by a line EF parallel to AB ; AD being 10 chains. 

6 a. 3 r. 12 p. = 68.25 sq. ch. = ABFE. 

15 a. = 150 sq. ch. = ABCD. 

A _ ABCD 150 sq. ch. % K _ 

AB = = = 15 ch. 

AD 10 ch. 

d „ ABFE 68.25 sq. ch. A ^ , 

AE = = = 4.5o ch. 

AB 15 ch. 

4. From a square ABCD, whose side is 9 chains, part off a triangle 
which shall contain 2 acres 1 rood 36 perches, by a line BE drawn from 
B to the side AD. 

2 a. 1 r. 36 p. = 24.75 sq. ch. = ABE. 

A _ 2 ABE 2 x 24.75 sq. ch. _ M . 

AE = = = 5.50 ch. 

AB 9 ch. 



308 SURVEYING. 






5. From ABCD representing the rectangle, whose length is 12.65 
chains, and breadth 7.58 chains, part off a trapezoid which shall contain 
7 acres 3 roods 24 perches, by a line BE drawn from B to the side DC. 



7 a. 3 r. 24 p. = 79 sq. ch. = ABED. 

ABCD = (12.65 x 7.58) sq. ch. = 95.887 sq. ch. 

A BCE = 95.887 sq. ch. - 79 sq. ch. = 16.887 sq. ch. 

„_ 2 BCE 2 x 16.887 sq. ch. / ■ _ 

CE = = = 4.456 ch. , nearly. 

BC 7.58 ch. J 

6. In the triangle ABC, AB = 12 chains, AC = 10 chains, and BC 
= 8 chains ; part off a trapezoid of 1 acre 2 roods 16 perches, by the line 
DE parallel to AB. 

1 a. 2 r. 16 p. = 16 sq. ch. = ABED. 

CAB= Vl5 x 3 x 5 x 7 sq. ch. = 39.6863 sq. ch. 
CDE = CAB - ABED = 39.6863 sq. ch. - 16 sq. ch. = 23.6863 sq. ch. 
CAB : CDE = ~CA 2 : ~CD 2 = CB 2 : CE 2 . 
39.6863 : 23.6863 = 10 2 : CD 2 = 8 2 : ~CE 2 . 

.-. CD = 7.725 ch. and CE = 6.18 ch. 
AD = CA - CD = 10 ch. - 7.725 ch. = 2.275 ch. 
BE = CB-CE= 8ch. -6.18 ch. = 1.82 ch. 

7. In the triangle ABC, AB = 26 chains, AC = 20 chains, and BC 
= 16 chains ; part off a trapezoid of 6 acres 1 rood 24 perches, by the 
line DE parallel to AB. 

6 a. 1 r. 24 p. = 64 sq. ch. = ABED. 

CAB = V31 x 5 x 11 X 15 sq. ch. = 159.9218 sq. ch. 
CDE = CAB - ABED = 159.9218 sq. ch. - 64 sq. ch. = 95.9218 sq. ch. 
CAB : CDE = ~CA 2 : CD 2 = ~CB 2 : ~CE 2 . 
159.9218 : 95.9218 = 20 2 : ~CD 2 = 16 2 : ~CE 2 . 

.: CD = 15.49 ch., and CE = 12.39 ch. 
AD = CA - CD = 20 ch. - 15.49 ch. = 4.51 ch., nearly. 
BE = CB - CE = 16 ch. - 12.39 ch. = 3.61 ch., nearly. 

8. It is required to divide the triangular field ABC among three 
persons whose claims are as the numbers 2, 3, and 5, so that they may 
all have the use of a watering place at C ; AB = 10 chains, AC = 6.85 
chains, and CB = 6. 10 chains. 

Since the triangles have the same altitude, they are to each other as 
their bases. Hence, it is necessary only to divide the base 10 into the 
three parts, 2 chains, 3 chains, 5 chains. 






teachers' edition. 309 

9. Divide the five-sided field ABCHE among three persons, X, Y, 
and Z, in proportion to their claims, X paying 3500, Y paying §750, 
and Z paying $1000, so that each may have the nse of an interior pond 
at P, the quality of the land being equal throughout. Given AB = 8.64 
chains, BC — 8.27 chains, CH = 8.06 chains, HE = 6.82 chains, and 
EA = 9.90 chains. The perpendicular PD upon A B = 5.60 chains, PJ7 
upon BC = 6.08 chains, PD" upon CH = 4.80 chains, .PIT" upon HE 
= 5.44 chains, and PD"" upon EA — 5.40 chains. Assume PH as the 
divisional fence between the shares of X and Z ; it is required to deter- 
mine the position of the fences PM and PN between the shares of X and 
Y, and between the shares of Y and Z. 

If P is joined to the vertices, the field is divided into triangles, whose 
bases are the sides, and the altitudes the given perpendiculars upon the 
sides from P. 

APB =8.64 x 2.80= 24.1920 sq. ch. 
BPC = 8.27 x 3.04 = 25.1408 
CPH = 8.06 x 2.40 = 19.3440 
HPE = 6.82 x 2.72 = 18.5504 
EPA = 9.90 x 2.70 = 26.7300 
ABCHE = 113.9572 sq. ch. 

The whole area, 113.9572 sq. ch., must be divided as the numbers 500, 
750, 1000, or as 2, 3, 4. 2 + 3 + 4 = 9. 

X's share = f of 113.9572 sq. ch. = 25.3238 sq. ch. 
Y's share = f of 113.9572 sq. ch. = 37.9857 sq. ch. 
Z's share = $ of 113.9572 sq. ch. = 50.6477 sq. ch. 

PH is assumed as the line between X's and Z's shares. Since the tri- 
angle PHE is less than X's share by 25.3238 sq. ch. - 18.5504 sq. ch. = 
6.7734 sq. ch. ; this difference must be taken from the triangle PEA 
The area of PEM is then 6.7734 sq. ch., and the altitude PIT" = 5.40 ch. 

., EM = ^^ = 2x 6.7734 sq.ch. = 2.5087 ch. 

PH"" 5.40 ch. 

PMA = PEA - PEM = 26.7300 sq. ch. - 6.7734 sq. ch. 
= 19.9566 sq. ch. 

Since Y's share is greater than PMA (19.9566 sq. ch.) and less than 
PMA + PAB (44.1486 sq. ch.), the point N is on AB. 
Y's share diminished by PMA equals PAN; that is, 

PAN= 37.9857 sq. ch. - 19.9566 sq. ch. = 18.0291 sq. ch. 

AN = 2 -^ = 2 X 18 -° 291 »* ch - = 6.4390 ch. 
PD 5.60 ch. 



310 SURVEYING. 

10. Divide the triangular field ABC, whose sides AB, AC, and BC 
are 15, 12, and 10 chains, respectively, into three equal parts, by fences 
EG and DF parallel to BC, without finding the area of the field. 

A ABC 



AAEG 



A ABC 
AADF 



3 AB 2 


AC 2 








2~z§T 2r 


"ztP* 




* 




3_ 225 
2"Z§~ 2 " 


.-. AE 2 = 


= 150. 


AE = Vl50 ch. 


= 12.247 ch 


3_ 144 
2~AG 2 ' 


.-. AG 2 -- 


= 96. 


AG= V96ch. 


= 9.798 ch. 


3 AB 2 


_AC 2 








i~z^ r 


' AF 2 








3 225 
1"Z5 2 ' 


.-. AD 2 = 


= 75. 


AD = V75 ch. 


= 8.660 ch. 


3 144 
1~AF 2 ' 


.-. AF 2 = 


= 48. 


^1F = V48 ch. 


= 6.928 ch. 



11. Divide the triangular field ABC, whose sides AB, BC, and AC are 
22, 17, and 15 chains, respectively, among three persons, A, B, and C, by 
fences parallel to the base AB, so that A may have 3 acres above the line 
AB, B 4 acres above A's share, and C the remainder. 



CAB = V27 x 5 x 10 x 12 sq. ch. = 127.2792 sq. ch. 
CDG = CAB - ABGD = 127.2792 sq. ch. - 30 sq. ch. = 97.2792 sq. ch. 
CEF= CAB - ABFE = 127.2792 sq. ch. - 70 sq. ch. = 57.2792 sq. ch. 
CAB : CDG = ~CB 2 : CG 2 = ~CA 2 : ~CD 2 . 
127.2792 : 97.2792 = 17 2 : ~CG 2 = 15 2 : CD 2 . 

.\ CG = 14.862 ch. and CD = 13.113 ch. 
CAB : CEF=~CB 2 : ~CF 2 = CA 2 : ~CE 2 . 
127.2792 : 57.2792 = 17 2 : OF 2 = 15 2 : CE 2 . 

.-. CF= 11.404 ch. and CE = 10.062 ch. 



Exercise IX. Page 295. 

1. Find the difference of level of two places from the following field 
notes: backsights, 5.2, 6.8, and 4.0; foresights, 8.1, 9.5, and 7.9. 

8.1 + 9.5 + 7.9 = 25.5 

5.2 + 6.8 + 4 = 16. 

9.5 



TEACHERS EDITION. 



311 



2. Stake of the following notes stands at the lowest point of a pond 
to be drained into a creek ; stake 10 stands at the edge of the bank, and 
10.25 at the bottom of the creek. Make a profile, draw the grade line 
through and 10.25, and fill out the columns H.G. and C, the former to 
show the height of grade line above the datum, and the latter, the depth 
of cut at the several stakes necessary to construct the drain. 



Station. 



+ S. 



H.I. 



-S.» 



H.S. 



H.G. 



Remarks. 



B 

1 
2 
3 
4 
5 



9 
10 
10.25 



6.000 



4.572 



10.2 

5.3 

4.6 

4.0 

6.8 

7.090 

3.9 

2.0 

4.9 

4.3 

4.5 
11.8 



25 



20.8 
20.4 
20.0 
19.6 
19.2 
18.8 
18.4 
18.0 
17.6 
17.2 
16.8 
16.7 



0.0 
5.3 
6.4 
7.4 
5.0 
5.1 
6.2 
8.5 
6.0 
7.0 
7.2 
0.0 



Bench on rook 
30 ft. west of 
stake 1. 




9 10 10.25 



5. Write the proper numbers in the third and fifth columns of the 
following table of field notes, and make a profile of the section. 



312 



SURVEYING. 



Station. 


+ S. 


H.I. 


-s. 


H.S. 


Remarks. 


B 


6.944 






20.0 


Bench on post 



1 




26.944 


7.4 
5.6 


19.5 
21.3 


22 ft. north 
of 0. 


2 






3.9 


23.0 




3 






4.6 


22.3 




t.p. 


3.855 




5.513 


21.431 




4 




25.286 


4.9 


20.4 




5 


.... 




3.5 


21.8 




6 






1.2 


24.1 






Exercise X. Page 301. 



1. The cross-section areas at five stations, 100 feet apart, of a railroad 
cut are, respectively, 576.8 square feet, 695.1 square feet, 809.5 square 
feet, 652.0 square feet, and 511.7 square feet. Compute the volume of 
material in this portion of the cut : (i) on the hypothesis that the cross 
sections are similar; (ii)'on the hypothesis that they are dissimilar, the 
alternate cross sections being regarded as mid sections. 

(i) V =iH{B+b+VEb). 

Vi = i x 100(576.8 + 695.1 + V576.8 x 695.1) ' 

= ^ (576.8 + 695.1 + 633.2) 

= ±P-x 1905.1 

= 63503.3. 
7 2 = Jx 100 (695.1 + 809.5 + V695.1 x 809.5) 

= i x 100(695.1 + 809.5 + 750.1) 

= ifo x 2254.7 

= 75156.7. 



TEACHEBS' EDITION. 



313 



Vz = i x 100(809.5 + 652.0 4- V809.5 x 652 0) 

= ix 100(809.5 + 652.0 + 726.5) 

= ^ x 2188 

= 72933.3. 
V 4 = i x 100(652.0 + 511.7 + V652.0 x 511.7) 

= i x 100(652.0 + 511.7 4-577.6) 

= H- x 1741.3 

= 58043.3. 

V = (63503.3 + 75156.7 + 72933.3 + 58043.3) cu ft 
= 269636.6 cu. ft. 
269636.6 

= — 5= — cu - y d - 



1.5 cu. yd. 
(ii) V =iH(B + b + 4M). 

Fi = i X 200(576.8 + 809.5 + 4 x 695.1) 
= 1 f il (576.8 + 809.5 + 2780.4) 

= if° x 4166.7 

= 138890. 
Y 2 = lx 200(809.5 + 511.7 + 4 x 652) 

= ^(809.5 + 511.7 + 2608) 

= ip x 3929.2 

= 130973.3. 
V = (138890 + 130973.3) cu. ft. = 269863.3 cu. ft. = 9994.9 cu. yd. 

2. Find the radius of a curve of 1°, of 2°. of 3°, of 4° of 5°. 
r = 50 esc i I). 



(i) r = 50 esc 0° 30'. 

log esc 0° 30' = 2.05916 
log 50 = 1.69897 
logr = 3.75813 

r = 5729.7 ft. 
(ii) r = 50 esc 1°. 

log esc 1° = 1.75814 

log 50 = 1.69897 

logr = 3.45711 

r = 2864.9 ft. 
(iii) r = 50 esc 1° 30'. 



logcscl°30' = 1.58208 
log 50= 1.69897 
logr = 3.28105 
r = 1910.1 ft. 
pv) r = 50 esc 2°. 

log esc 2° = 1.45718 
log 50 = 1.69897 
logr = 3.15615 
r = 1432.7 ft. 
(v) r = 50 esc 2° 30'. 

log esc 2° 30' = 1.36032 
log 50 = 1.69897 
logr = 3.05929 
r = 1146.3 ft. 



314 



SURVEYING. 



3. Two adjacent straight sections of a railroad form an angle of 148° 
16". They are joined by a curve touching each of them at the distance 
of 388 feet from the vertical point. Find the radius and the degree of 
the curve. 



r = t tan -J- A . 
A = 148° 16'. 


50 
sin I D = — cot i A. 
t 


.-. $ A = 74° 8'. 


log 50 = 1.69897 


t = 388 ft. 


colog 388 = 7.41117 -10 


log 388= 2.58883 


log cot 74° 8' = 9.45367 


tan 74° 8' = 10.54633 


logsin£D = 8.56381 


logr = 3.13516 


iD = 2°5 / 56.7 // 


r = 1365.1 ft. 


B = 4° 11' 53". 



NAVIGATION. 



Exercise I. Page 319. 

1. Given compass course S., wind E.8.E., leeway 1J points, variation 
52° 0' W., deviation 2° 0' E. ; required true course. 

Since the wind is E.S.E., and the compass course is S., the ship is on 
the port tack ; hence, leeway is allowed to the right. 

Compass course pts. R. of S. 

Leeway ljr pts. R. 

Compass course corrected for leeway . 1J pts. R. of S. 

= 14° 3' 45" R. of S. 
Variation and deviation (52° — 2°) W. = 50° 0' 0" L. 

35° m' 15" L. of S. 
True course, S. 35° 56' E. 

2. Given compass course W.N.W., wind N., leeway 3 points, variation 
42° 0' E., deviation 18° 30' W7] required true course. 

Since the wind is N., and the compass course is W.N.W., the ship is on 
the starboard tack ; hence, leeway is allowed to the left. 

Compass course pts. L. of N. 

Leeway 3 pts. L. 

Compass course corrected for leeway . 9 pts. L. of N. 

= 7 pts. R. of S. 
= 78° 45' R. of S. 
Variation and deviation (42° - 18° 30') E. = 23° 30' R. 

102° 15' R. of S. 
77° 45' L. of N. 
True course, N. 77° 45' W. 

3. Given compass course S.S.E. \ E., wind S. W. -J- S., leeway &J points, 
variation 2i points E., deviation 1£ points W. ; required true course. 

Compass course 2-J- pts. L. of S. 

Leeway (starboard tack) .... 3-^- pts. L. 

Variation and deviation .... 1 pt. R. 

5 pts. L. of S. 

True course, S.E. by E. 
* 315 



316 NAVIGATION. 

4. Given true course S. 79° W. , wind S. by W. , leeway f point, varia- 
tion 10° 30' E., deviation 19° 0' W. ; required compass course. 

True course . . , . . . 79° R. of S. 

Leeway (port tack) .... 8° 26' 15" L. 

Variation and deviation . . . 8° 30' R. 

79° 3' 45" R. of S. 
Compass course, S. 79° 4' W. 

5. Given compass course W. £ N., wind N.N.W., leeway If points, 
variation 8° 30' E., deviation 15° 35' E., required true course. 

Compass course . . . . . 7f pts. L. of N. 
Leeway (starboard tack) . . . If pts. L. 



9i pts. 




L. 


of N. 


= 6* pts. 




R. 


of S. 


= 73° r 


30" R. 


of S. 


24° 5' 




R. 





Variation and deviation 

97° 12' 30" R. of S. 

= 82° 47' 30" L. of N. 
True course, N. 82° 47' W. , 

6. Given compass course E. JN., wind N.N.E., leeway 2J- points, vari- 
ation 13° / W., deviation 20° 0' E.; required true course. 

Compass course 7f pts. R. of N. 

Leeway (port tack) 2j pts. R. 

10 pts. R. of N. 

= 6 pts. L. of S. 

Variation and deviation .... = 67° 30' L. of S. 

7° R. 



60° 30' L. of S. 
True course, S. 60° 30' E. 

7. Given true course S. 85° E., wind N. by W., leeway -J- point, vari- 
ation 14° 0' E., deviation 19° 0' E.; required compass course. 

True course 85° L. of S. 

Leeway (port tack) .... 5° 37" 30" L. 

Variation and deviation . . . 33° L. 

123° 37' 30" L. of S. 
= 56° 22' 30" R. of N. 
Compass course, N. 56° 22' E. 

8. Given compass course W., wind N.N. W., leeway 1£ points, variation 
18° 30' E., deviation 21° 0' W. ; required true course. * 



5 


pts. 


L. 


of S. 


= 56 


°15 y 


L. 


of S. 


17 


o 


R. 





teachers' edition. 317 

Compass course 8 pts. L. of N. 

Leeway (starboard tack) .... lj pts. L. 

9i pts. L. of N. 
= 6f pts. R. of S. 
= 75° 56' 15" R. of S. 

Variation and deviation . . . 2° 30" L. 

73° 26' 15" R. of S. 
True course, S. 73° 26' W. 

9. Given compass course E. }S., wind X.N.E. £E., leeway 2£ points, 
variation 21° 0' E., deviation 4° 0' W. ; required true course. 

Compass course 7i pts. L. of S. 

Leeway (port tack) 2j pts. R. 



Variation and deviation 

39° 15' L. of S. 
True course, S. 39° 15' E. 

10. Given true course, E. by S. i S., wind X. by W., leeway 2f points, 
variation 2 points W., deviation 3-J- points E.; required compass course. 

True course . . . . * . . . 6£ pts. L. of S. 

Leeway (port tack) 2f pts. L. 

Variation and deviation .... lj pts. L. 

11 pts. L. of S. 
= 5 pts. R.of N. 
Compass course, N.E. by E. 

11. Given true course N. by W., wind N.E., leeway 3J points, variation 
2f points E., deviation 1^ points E.; required compass course. 

True course 1 pt. L. of N. 

Leeway (starboard tack) 3^ pts. R. 

Variation and deviation .... 4j- pts. L. 

2 pts. L. of N. 
Compass course, y/.N.AV. 

12. Given true course N.N.W., wind S.S.W., leeway 2 -J- points, vari- 
ation 2f points E., deviation £ point E. ; required compass course. 

True course 2 pts. L. of N. 

Leeway (port tack) 2\ pts. L. 

Variation and deviation .... Sj pts. L. 

8 pts. L. of N. 
Compass course, W. 



318 NAVIGATION. 

13. Given true course S. 64° E., leeway 0, variation 7° 0' W., devia- 
tion 15° 0' W. ; required compass course. 

True course ' 64° L. of S. 

Variation and deviation 22° R. 

42° L. of S. 

Compass course, S. 42° E. 

14. Given true course N. 44 W., leeway 0, variation 6° 0' E., deviation 
20° 0' W. ; required compass course. 

True course . 44° L. of N. 

Variation and deviation* 14° R. 

30° L. of "K 
Compass course, N. 30° W. 

15. Given compass course N. 65° W., leeway 0, variation 10° 0' E., 
deviation 3° 0' E. ; required true course. 

Compass course 65° L. of N. 

Variation and deviation 13° R. 

52° L. of N. 
True course, N. 52° W. 

16. Given compass course S. 15° W., leeway 0, variation 0° 0' W. s 
deviation 18° 0' E. ; required true course. 

Compass course 15° R. of S. 

Variation and deviation 12° R. 

27° R. of S^ 
True course, S. 27° W. 

17. Given compass course S. 18° E., leeway 0, variation 25° 0" E., 
deviation 10° 0' E., required true course. 

Compass course 18° L. of S. 

Variation and deviation 35° R. 

17° R. of S. 
True course, S. 17° W. 

18. Given compass course N. 30° E., wind S. by W., leeway 1£ points, 
variation 12° 0' E., deviation 10° 0' W. ; required true course. 

Compass course 30° R. of N. 

Leeway (starboard tack) . . . 14° 3' 45" L. 
Variation and deviation . . . 2° R. 

17° 56' 15" R. of N. 
True course, N. 17° 56" E. 



teachers' edition. 



319 



Exercise II. Page 332. 
1. Given L' 49° 57' N., C S. W. by W., I) 488.0 ; required L" and p. 



Z>= 488.0 
C = 56° 15' 



p — D sin C. 

log D = 2.68842 
log sin = 9.91985 
logp = 2.60826 
p = 405.8. 



2. Given 1/ 1° 45' N., C S.E. by E., D 487.8 



J) =487.8 
C = 56° 15' 



p = D sin C. 

logD= 2.68824 

log sin C = 9.91985 

logp = 2.60809 

p = 405.6. 



L d — I) cos C. 

log D = 2.68842 

log cos C = 9.74474 

log i d = 2.43316 

X d = 271' 

= 4°31'S. 
2y= 49°57 / N. 
L // =45°26 / N. 

required If and p. 

La— D cos C. 

logD = 2.68824 

log cos C = 9.74474 

logi d = 2.43298 

L d = 271' 

= 4° 31' S. 
X / = l°45 / N. 
L" = 2° 46' S. 



3. Given L' 3° 15' S., C N.E. by E. f E., D 449.1 ; required L" and p. 



D = 449.1 

C = 64° 41' 15" 



p = D sin C. 

log 2) =2.65234 
log sin C = 9.95616 
logp = 2.60850 
p = 406. 



Ld — B cos C. 

logi) = 2.65234 

log cos C = 9.63099 

logi d = 2.28333 

L d = 192' 

= 3° 12' N. 
Z,:=3°15' S. 



Z" = 0° 3' S. 



4. Given U 2° 10' S., C N. by E., D 267.0 ; required L" and p. 



D= 267.0 
C=ll°15 / 



p — T> sin C. 

logD= 2.42651 

log sin C = 9.29024 

logp= 1.71675 

p = 52.1 



X^ = D cos C. 

log!) =2.42651 

log cos C = 9.99157 

logX d = 2.41808 

L d = 262 / 

= 4°22'N. 
1/ = 2° 10 / S. 
X // = 2°12 / N. 



320 



NAVIGATION. 



5. Given i' 41° 30' N., CS.S.W.,D 295.5 ; required L" and p. 



B- 
C = 



= 295.5 
: 22° 30' 



p = D sin 0. 

log D= 2.47056 

log sin = 9.58284 

log p = 2.05340 

p = 113.1. 



Ld — B cos O. 

log i = 2.47056 

log cos C = 9.96562 

log id = 2.43618 

id = 273' 

= 4°33 / S. 
i'=4J/WN. 
2/'= 36° 67' N. 



6. Given i' 21° 59' S., i" 24° 49' S., i 360 ; required O and p. 



B =360 
£<* =170 



cos O = -=j • 

log id = 2.23045 

logD = 2.55630 

log cos 0=9.67415 

C = 61° 49' 

= 5-J- pts. , nearly. 



pa=(D-i d )(D+i d ) 
= 190 x 530. 

log 190 = 2.27875 

log 530 = 2.72428 

log p 2 = 5.00303 

logp = 2.50151 

p = 317.3. 



7. Given i' 2° 9' S., i" 3° ll 7 N., i 354 ; required O and p. 



B =354 
id =320 



cosO = ^- 

log id = 2.50515 

log D= 2.54900 

log cos 0=9.95615 

O = 25° 19' 

= 2J pts. 



p 2 



> 2 =(i-id)(i+id) 
= 34 x 674. 

log 34 = 1.53148 

log 674 = 2.82866 

log p2- 4.36014 

logp = 2.18007 

p = 151.4. 



8. Given i' 1° 30' N., i" 0° 26' S.,CS. by W. ; required i and p. 



L d = 116 
O = 11° 15' 



D = La sec O. 

log i d = 2.06446 

log sec O = 0.00843 

log!) = 2.07289 

i= 118.3. 



p — Ld tan O. 
log i d = 2.06466 
log tan 0= 9.29866 
logp= 1.36312 
p = 23.1. 



9. Given i' 40° 17' N. , L" 37° 6' N. , O S. by W.{W.; required B and p. 



id =191 

O =16° 52' 30' 



B — Ld sec O. 
log id = 2.28103 
log sec 0= 0.01911 
log B= 2.30014 
i= 199.6. 



p — Ld tan O. 
log id = 2.28103 
log tan 0= 9.48194 
logp= 1.76297 
p = 57.9. 



TEACHERS EDITIOX. 



321 



10. Given U 38° 0' N., C S. W. by W., p 48.2 ; required 2/' and D. 



p = 48.2 
C = 56° 15' 



B — p esc C. 

log p = 1.68305 

log esc C = 0.08015 

log D= 1.76320 
D=58.0. 



: p COt C. 

■ 1.68305 
log cot = 9.82489 



L d 

l0gp: 



logi d = 1.50794 

£<* = 32' S. 

2/ = 38° 0'N. 

I/ / = 37 28 / N. 



11. Given L' 18° 25' N., C S.W. by W. fW., p 65.1 ; required L" 



and D. 



p = 65.1 

O = 64° 41' 15" 



D = p esc C. 

logp = 1.81358 

log esc C = 0.04384 

logZ> = 1.85742 
Z>= 72.02 



L d —p cot C. 

logp= 1.81358 

log cot C = 9.67483 

logZ, d = 1.48841 
X rf = 31' S. 
Z/ = 18°25'N. 



L // = 17°54 / N. 



12. Given £' 50° 18' N., X" 54° 48' N., D 299.0 ; required C and p. 



D = 299.0 
L d = 270 



cosC = ^- 

log£ d = 2.43136 
log D= 2.47567 



log cos C = 9.95569 

C = 25° 26' 30" 
= 2Jpts., nearly. 



p*=(D-L d )(I)+L d ) 
= 29 X 569. 
log 29= 1.46240 
log 569 = 2.75511 



logp2 = 4.21751 
logp = 2.10875 
p = 128.5. 



13. Given 2/ 32° 30' N., L" 19° 59' N., D 812.0 ; required C and p. 



D = 812.0 
X d = 751 



cosC = ^« 

logZ, d = 2.87564 
logD = 2.90956 

log cos (7=9.96608 
C = 22° 21' 
= 2 pts. , nearly. 



&=(D-L d )(D+L d ) 
= 61 X 1563. 

log 61= 1.78533 
log 1563 = 3.19396 

logp 2 = 4.97929 
logi> = 2.48964 
p = 308.8. 



322 



NAVIGATION. 



14. Given L' 2° 8' S., C N. 11° E., I) 500 ; 



B 



500 
11° 



p — B sin C. 

logD = 2.69897 
log sin C = 9.28060 
logp = 1.97957 
p = 95.4. 



required L" and p. 

X rf = B cos C 
log I>= 2.69897 
log cos C = 9.99195 
\ogL d = 2.69092 
X rf = 491' 

- 8° 11' N. 
1/ - 2° 8 / S. 
£" = 6° 3'N. 



15. Given 2/ 20° 21' S., C N. 20° E., D 402.0 ; required Z," and p. 



D= 402.0 
C = 20° 



p = D sin C. 

log D= 2.60423 

log sin C = 9.53405 

logp = 2.13828 

p =137.5. 



La — D cos C 

logD = 2.60423 
log cos C = 9.97299 
logZ d = 2.57722 
L d = 378' 

= 6° 18' N. 
X' = 20° 21' S. 
X" = 14° 3' S. 



16. Given X' 40° 25' S., C N. 87° E., I) 240.0 ; required X" and p. 



D= 240.0 

C=87° 



p -■ B sin C. 

logD = 2.38021 
log sin C - 9.99940 
logp = 2.37961 
p = 239.7. 



La — I> cos C. 
logD = 2.38021 
log cos C = 8.71880 
log i d = 1.09901 
L d = 13' N. 
U = 40° 25' S. 
I/' = 40° 12' S. 



17. Given If 20° 48' N., L" 17° 13' N., p 289.2 W. ; required C and D. 



p = 289.2 
L d ^2l5 



tan C - -£- • 

logp= 2.46120 

logX rf = 2.33244 

log tan 0=10.12876 

O = S. 53° 22' 18" 

= S. 53° 22' W. 



W. 



B — p esc C. 

logp = 2.46120 

log esc C - 0.09554 

logD= 2.55674 

B= 360.4. 



TEACHEKS' EDITION. 



323 



18. Given 2/ 51° 45' N., L" 53° 11' X., p 128.0 E. ; required C and 1). 



p= 128.0 . „ p 

x d =86 tan6 = x:/ 

logp = 2.10721 

\ogL d = 1.93450 

log tan C = 10.17271 

C = N. 56° 6' 13" E. 
= N.E. by E. nearly. 



I) — p cse C. 

logp= 2.10721 

log esc C — 0.08090 

log 1) = 2.18811 

2> = 154.2. 



19. Given 2/ 0° 20 / S., If 0° 18' N., p 142.7 E. ; required C and D. 



p — 142.7 nC= P - 

La — 38 La 

logp= 2.15442 

logX^= 1.57978 

log tan = 10.57464 

C = N. 75°5 / 19 // E. 
= N. 75°5 / E. 



]J — p esc C. 

logp = 2.15442 

log esc C = 0.01488 

logZ> = 2.16930 

2)= 147.7. 



20. Given L' 40° 20 / N., L" 41° 37' N., p 52.6 W. ; required C and D. 



p = 52.C 
L d =17 



tan C = #- 



logi>= 1.72099 

log i d = 1.88649 

log tan C = 9.83450 

C=N.34°20'17"W. 
= N.34°20 , W. 



D = p esc C. 

logp= 1.72099 

log esc C = 0.24867 

log£> = 1.96966 

Z) = 93.3. 



Exercise III. Page 335. 

1. Given L 55° 55', X 7 2° 10' W., X" 12° 52' E.; required p. 

p = lo-i- x 33.62 = 505.4 E. 

2. Given X 52° 0', Y 0° 59' W., X" 2° 24' E.; required p. 

p = 3|s x 36.94= 125.0 E. 

3. Given i 61° 25', V 179° 20' W., X" 176° 52' E.; required p. 

p = 3|8 x 28.71 = 109.1 W. 

4. Given L 56° 0', V 3° 12' W., X" 4° 8' E. ; required p. 

p-l^X 33.55 = 246.0 E. 



324 NAVIGATION. 

•*• 
5- Given L 80° 0', X' 10° 0' W., X" 17° 41 / W.; required p. 

p = 7|JX 10.42 = 80.1 W. 

6. Given L 60° 0', p 204.0 E., X 7 160° 2' E. ; required X". 

\ d = 204.0 -f i = 408' = 6° 48' E. ; 
X" = 160° 2' + 6° 48' = 166° 50' E. 

7. Given X 51° 28', p 70.9 E., X' 32° 7' W.; required X". 

Xd = 70.9 -f 37.38 = 1.90° = 1° 54' E.; 
\" = 32° T — 1° 54 / = 30° 13' W. 

8. Given L 64° 16', p. 265.7 W., X' 170° 0' W.; required X". 

X d = 265.7 -f 26.05 = 10.20 = 10° 12' W. ; 

V = 170° 0' + 10° 12' = 180° 12' W. = 179° 48' E. 

9. Given L 46° 37', p 352.0 E., X / 163° 42 / E. ; required X". 

\ d = 352.0 -f 41.21 = 8.54° = 8° 33' E. ; 
\" = 163° 42' + 8° 33' = 172° 15' E. 

10. Given L 39° 57', p 398.0 W., V 4° 8' W. ; required X". 

\ d = 398.0 4- 45.93 = 8.67° = 8° 39' W. ; 

y, _ 4 o g, + go 39/ = 12 47/ W< 

11. From latitude 32° 3' S., longitude 179° 45' W., a ship makes 54 
miles west (true). Required the longitude in. 

\ d = 54 -f 50.85 = 1.06° = 1° 4' W. ; 

X" = 179° 45' + 1° 4' = 180° 49 / W. = 179° 11 / E. 

12. From latitude 35° 30' S., longitude 27° 28' W., a ship sails east 
(true) 301 miles. Required the longitude in and the compass course; 
variation If points E., leeway £ point to the left, deviation 8° 50' E. 

X d = 301 -f 48.85 = 6.16° = 6° 10' ; 
X" = 27° 28' - 6° 10' = 21° 18' W. 

True course '8 pts. R. of N. 

Variation and leeway 2 pts. L. 

6 pts. R. of N. 
= 67° 30' R. of N. 

Deviation . 8° 50' L. 

Compass course . . . . . . N. 58° 40' E. 



teachers' edition. 



325 



Exercise IV. Page 340. 



1. Given L' 25° 35 
required C and D. 
L d = 1° 53' 

= 113' 
L m = 26° 32' 
\ d = 5° 5' 
= 305' 



N., L" 27° 28' X., X' 60° 0' W., X" 54° 55' W.; 



tan C = 



log \ d = 

logcos£ m = 

cologZd == 



Xd cos Z, m 

L d 
2.48430 
9.95167 
7.94692 



log tan C- 10.38289 
C=N. 67°30'E.= E.X.E. 



B— Ld sec C. 

logi d = 2.05308 
log sec C = 10.41716 
logD = 2.47024 
Z>= 295.3. 



2. Given 2/ 32° 30' N., 7," 34° 10' N., X' 25° 24' W., X" 29° 8' W.; 



required C and D. 
L d = 1° 40' 

= 100' 
L m = 33° 20' 
\ d = 3° 44' 

= 224' 



tanC: 



Xd cosX^ 



id 

logXd = 2.35025 

logcosi m = 9.92194 

cologX rf = 8.00000 

log tan C = 10.27219 

.-. C = N. 61° 53' W. 



D = Ld sec C. 

log id = 2.00000 

log sec C = 10.32673 

logD= 2.32673 

i> = 212.2. 



3. Given i' 39° 30' S., X" 41° 0' S., X' 74° 20' E., X" 70° 12' E. ; required 



2) = Ld sec C. 

log id = 1.95424 

log sec C = 10.36708 

logi> = 2.32132 

i> = 209.6. 



4. Given i' 46° 24' S., X' 178° 28' E., C S.E. f E., D 278.0; required 
/' and X". 
C = 53° 26' 
D = 278.0 



(7 and i>. 




i d = 1° 30' 
= 90' 


Xd COS i m 

tan = 


L m = 40° 15' 


logXtf = 2.39445 


\ d =4° 8' 


logcosi,„ = 9.88266 


= 248' 


colog id = 8.04576 




log tan C = 10.32287 




... C = S. 64° 34' W. 



Ld — I) cos C. 
log D= 2.44404 
log cos C = 9.77507 
logLd =2.21911 
L d = 166' 

= 2° 46' 
i' = 46° 24' 
L" = 49° 10' S. 
i m = 47°47'. 



Xd = D sin C sec L m . 
logD = 2.44404 
log sin C = 9.90480 
log sec i m = 10.17267 



logXd 


= 2.52151 


Xd 


= 332' 






= 5° 


32' 


X' 


= 178° 


28' 


X" 


= 184° 


0'E. 




= 176° 


0' W, 



326 



NAVIGATION. 



5. Given L' 20° 29' N., X' 179° 10' W., C W. by S. ± S., B 333.0; 



required L" and X". 
C = 73° 7 
B = 333 



La — B cos 0. 
log 2) = 2.52244 
log cos (7=9.46303 

log£ d = 1.98548 
^ = 97' 

= 1°37 / 
2/ = 20° 29' 



Z"=18°52'N. 
L m = 19° 40'. 



\d — B sin (7 sec _L m . 
log 2) = 2.52244 
log sin C= 9.98087 
logsec£ m = 10.02610 

logX^= 2.52941 

X d = 338' 

= 5° 38' 
\' - 1790 10' 



X- ±= 184° 48' W. 
= 175° 12' E. 



6. Given 1/ 0° 56' N., V 29° 50' W., C S. 47° E., D 168.0 ; required 

." and X". 
C = 47° 
D=168 



La — B cos C. 

logD = 2.22530 

log cos C- 9.83378 

logX d = 2.05908 
L d = 115' 

= 1° 55' 
2/ = o° 56' 



I" = 0° 59' S. 
£m = 0° 1'. 



\d = B sin C sec X^ 
log 7) =2.22530 
log sin C = 9.86413 
log sec L m = 0.00000 

log \ d = 2.08943 
\ d = 123' 

= 2° 3' 
X' = 29° 50' 



X" = 27° 47' W. 



7. Given X' 42° 25' N., X' 66° 14' W., C S.E. by E., B 25.0; required 



' and X". 

C = 56° 15' 
D=25.0 



Ld — B cos C 

log D= 1.39794 

log cos 0=9.74474 

logL d = 1.14268 
Xd= 14' 

£' = 42° 25' 



£" = 42° 11' N. 
i m = 42° 18'. 



Xd = D sin O sec L m . 
logD= 1.39794 
log sin C = 9.91985 
log seci m = 0.13098 

log Xd= 1.44877 

\ d = 28' 
X' = 66° 14' 



X" = 65° 46' W. 



8. Given £' 42° 8' N., X' 65° 48' W., C E. *S., B 126.0; required 
L" and X". 



TEACHERS' EDITION. 



327 



C = 84° 22 / 

D = 126.0 



L d — B cos C. 
logD= 2.10037 
log cos C = 8.99194 

logi rf = 1.09231 
£d = 12' 
Z/ = 42° 8' 



2/' = 41° 56' N. 

Z m = 42° 2'. 



Xd = D sin C sec L n 
log£>= 2.10037 
log sin C = 9.99790 
log sec L m = 0.12915 

logi^ = 2.22743 
X d = 168' 

= 2°48 / 
V = 65° 48' 



X" = 63° 0' W. 



9. Given 2/ 41° 52' N., V 62° 47' W., CE.} S., D 161.0; required 
L" and X 7 '. 



C = 84° 22' 
B = 161.0 



La — B cos C 

log D = 2.20683 

log cos C=8.99194 

logi d = 1.19877 
L d = 16' 

L' = 41° 52' 



£" = 41° 36' N. 
L m = 41° 44' N. 



\a — B sin C sec Z Wi 
log D= 2.20683 
log sin C = 9.99789 
log sec i m = 0.12712 



logX rf = 2.33184 
Xd = 215' 

= 3°35 / 
X 7 = 62° 47' 



X" = 59° 12' W. 



10. Given L' 41° 38' N., L" 41° 26' N., V 59° 16' W., C E. by S.; 



required X" and B. 



I<d =12 

Z m = 41°32' 

C = 78° 45' 



D = id sec C. 

log id= 1.07918 

log sec G = 0.70976 



logD = 1.78894 
Z>=61.5. 



Xd = X) sin C sec £„ 
logD = 1.78894 
log sin C = 9.99157 
log sec L m — 0.12577 



logXd= 1.90629 
Xd = 81' 

= 1°21' 
V = 59° 16' 



X" = 57° ory W. 



11. Given L' 41° 19' N., L" 41° 11' N., V 57° 47' W., B 167.0 ; required 
X" and C 



328 



NAVIGATION. 



L m = 41° 15' 
D= 167.0 



cos C 



B 

logi d = 0.90309 

log B- 2.22272 

log cos C = 8.68037 

C = 87° 15'. 



\d = B sin C sec L w . 

logD = 2.22272 

log sin C - 9.99950 

log sec L m = 0.12387 

log\d = 2.34609 

Xd = 222' 

= 3° 42' 
V = 570 47/ W# 



X" = 61° 29' W. 
or 54° 5'W. 



12. Given X' 46° 28' N., Z" 45° 17' N., X 7 22° 18' W., X" 19° 39' W.; 



required C and D. 

£4 = 71 

Z w = 45° 52' 
Xd = 159 



tan C — 

log X d = 

log cos L m — 

colog£ d = 



\d cos X w 

2.20140 

9.84282 
8.14874 



log tan C = 10.19296 

C = S. 57° 19' E. 



B — La sec C. 
logZ d = 1.85126 
log sec C - 0.26761 
logD = 2.11887 
D= 131.5. 



13. Given 1/ 25° 30' S., Z" 28° 15' S., X' 2° 15' E., X" 11° 17' E. 



required C and B. 

L d = 165 
X w = 26° 52' 
X d = 542 



tan0= x^x^ 

log\ d = 2.73399 

logcos_L m = 9.95039 

cologX d = 7.78252 

log tan C - 10.46690 

C = S. 71° 9' E. 



B— Ld sec C. 
logX d = 2.21748 
log sec C = 0.49067 
log B =2.70815 
B= 510.7. 



14. Given L' 33° 40' N., X" 30° 49' N., X' 13° 20' E., X" 17° 56' E.; 



required C and Z>. 

Zd = 171 
L m = 31° 44 
\d = 276 



tan = 



\d cos i^ 



Zd 

logX<* = 2.44090 

logcosZ w = 9.92968 

cologZd = 7.76700 

log tan (7=10.13758 

C = S. 53° 46' E. 



B— Ld sec C 
\ogL d = 2.23300 
log sec C = 0.22836 
logZ> = 2.46136 
B = 289.3. 



teachers' edition. 



329 



15. Given L' 19° 30 / S., L" 17° 24' S., X' 0° 10' E., X" 1° 28' W.; 



required C and D. 
L d = 126 
L m = 18° 27' 
Xd = 98 



tanC = 



X c / cos L m 



L d 

log x d = 1.99123 

log cos£ m = 9.97708 

cologZ d = 7.89963 

log tan C- 9.86794 

C=N.36°25'W. 



B — La sec C. 
logL d = 2.10037 
log sec C = 0.09435 
logD= 2.19472 
D= 156.6. 



16. A ship sails from Boston light-house, in latitude 42° 20' N., longi- 
tude 71° 4' W., on a N.N.E. course, 184 miles. Find the latitude and 
longitude in. 

C = 22° 30' 

D=184 



La = L> cos C. 

logD= 2.26482 
log cos C = 9.96603 
logi d = 2.23085 
L d = 170 / 

= 2°50 / 
L' = 42° 20' 



L" = 45° 10' N. 
i m = 43° 45'. 



Xd = D sin C sec Z„ 

logD= 2.26482 

log sin C = 9.58284 

log sec L m - 0.14124 

log\ d = 1.98890 

Xd = 97' 

= 1°37 / 
V = 71° 4 / 
X" = 69° 27' W. 



17. A ship sails from Cape May, in latitude 38° 56' N., longitude 



74° 57' W., on a S.S.E. course, 240 miles, 
tude in. 

C = 22° 30' L d = D cos C. 

D=240 logD = 2.38021 

log cos C = 9.96603 
\ogL d = 2.34624 
X d = 222' 

= 3° 42' 
jj = 38° 56' 



35° 14' 
£ m = 37° 5 / 



N. 



Find the latitude and longi- 

\d — T> sin C sec L m . 

logD = 2.38021 

log sin C = 9.58284 

log sec L m - 0.09813 

logX rf = 2.06118 

Xd = 115" 

= i°55' 
\' = 74° 57' 
X"=73° 2'W. 



18. A ship sails from Cape Cod light, in latitude 42° 2' N. , longitude 
70° 3" W., on an E. by N. compass course, 170 miles; wind S.E. by S., 
leeway i point, deviation 17f° E., variation llj W. Find the latitude 
and longitude in. 



330 



NAVIGATION. 



Compass course 
Leeway 



Variation and deviation 



7 pts. R. of N. 
ipt. L. 



6f pts. 


R. 


of N. 


= 75° 66' 


R. 


of N. 


6°30 / 


R. 





True course N. 82° 26' E. 



O = 82° 26' 
D=170 



La — B cos 0. 

logD = 2.23045 

log cos = 9.11951 

log i d = 1.34996 

id = 22' 

L' = 42° 2' 



L" = 42° 24' N. 
L m = 42° 13'. 



Xd = B sin O sec L n 

logD = 2.23045 

log sin = 9.99620 

log sec L m = 4 13041 

log Ad = 2.35706 

Xd = 228' 

= 3° 48' 
V= 70° 3" 
X- = ffio 15 / w> 



19. A ship sails from Cape Cod light on a S.S.E. compass course, 140 
miles; deviation 5^° E., variation 11£° W. Find the latitude and longi- 
tude in. 

Compass course 22° 30' L. of S. 

Variation and deviation .... 5° 45' L. 

True course S. 28° 15' E. 



C = 28° 15' 
D=140 



id = D cos O. 

log D = 2.14613 

log cos C = 9.94492 

log id = 2.09105 

L d = 123' 

= 2°3 / 
i' = 42° 2' 



L" = 39° 59' N. 
i w = 41°0'. , 



Xd = D sin O sec i m . 

log!) =2.14613 

log sin 0=9.67516 

log sec L m - 0.12222 

log Xd= 1.94351 

Xd = 88' 

= 1°28 / 
X' = 70° 3' 



° 35' W. 



20. A ship sails from latitude 55° V N., longitude 1°25' W., on a S.W. 
compass course, 101 miles; wind W.N. W., leeway 1 J points, deviation 
6° W. , variation 24° 56' W. Find the latitude and longitude in. 

Compass course ... 4 pts. R. of S. 

Leeway lj pts. L. 

2f pts. R. of S. 
= 30° 56' R. of S. 
Variation and deviation 30° 56' L. 
True course .... S. 



L d 


= D 




Xd 


= 0. 






= 101 




\" 


= X' 






= 1° 


4r 




= 1° 25' 


W. 


L' 


= 55° 


V 








L" 


= 53° 


20' N. 







TEACHERS EDITION. 



331 



21. A ship sails from the Bermudas, in latitude 32° 18' N., longitude 
64° 50' W., on a W.S.W. compavss course, 190 miles; deviation 1 point 
W. , variation 1 point W. Find the latitude and longitude in. 



Compass cou 
Variation an 


rse 

I deviation 

La — D cos C. 

log D= 2.27875 

log cos C = 9.84949 


log 


6 pts. R. of S. 
2 pts. L. 


True course 

C =5 45° 
D =190 


4 pts. R. of S. 

\d — D sin C sec L 
log£ = 2.27875 
5 sin C= 9.84949 




logX d = 2.12824 
L d = 134' 

= 2° 14' 
V = 32° 18' 


secX m = 0.06777 




log\ d = 2.19601 
\ d = 157' 
= 2° 37' 




L" = 30° 4'N. 

X w = 31° 11'. 


\' = 64° 50' 




X" = 67° 27' W. 



22. A ship sails from, the Bermudas on a W.N.W. compass course, 90 
miles ; wind S.W., leeway 1 point, deviation 1 point E., variation 1 point 
W. Find the latitude and longitude in. 

Compass course . 

Leeway 

Variation and deviation 

True course . 

C = 56° 15' 
D = 90 






6 pts. L. 
1 pt. R. 



of N. 


5 pts. L. 
56° 15' L. 


of N. 
of N. 



Ld — D cos C. 

log D= 1.95124 

log cos = 9.74474 



log£tf=l.C 

L d = 50' 
X' = 32° 18' 



jr/^33 8'N. 
L m = 32° 47'. 



\ d — D sin C sec L ni 
logD= 1.95124 
log sin C = 9.91985 
log sec L m = 0.07502 

logX d = 1.94911 
\ d = 89' 

= 1°29' 
\' = 64° 50' 



\" = 6Qo 19 / w# 



23. A navigator wishes to sail on a rhumb from the Bermudas to Cape 
Fear, in latitude 33° 52' N., longitude 78° W. ; variation 10° W., deviation 
7° W. Find the compass course and distance. 






332 



NAVIGATION. 



£<* = 94 

L m = 32° 5' 
\ d = 790 



tan C ' = 



Xd cos £„, 



logXd = 2.89763 

logcosZ w = 9.92318 

cologi d = 8.02687 

log tan C = 10.84768 

C=N. 81°55'W. 
Var. anddev. = 17° W. 



Compass course N. 64° 55' W. 



D = Ld sec C 
log L d =1.97313 
log sec C = 0.85197 
logD = 2.82510 
D = 668.5. 



24. A ship from latitude 36° 32' N. sails between south and west until 
she has made 480 miles of departure, and 9° 22' of difference of longi- 
tude. Kequired the latitude in, the course steered, and the distance run. 
[Take L m = i (2/ + L") + 13'.] 



p = 480 
\ d = 562 


I T P 

cos£ m = f-- 

Xrf 


tanC= 2— 
Ld 




log p = 2.68124 


logp = 2.68124 




log X d = 2.74974 


logL d = 2.81291 




log cos£ w = 9.93150 


log tan 0=9.86833 




L m = 31° 20'. 


O = S. 36° 27' W 


* 


£" = 2(i w -13')-Z,' 

= 25° 42' N. 
L d = 10° 50' 

= 650. 


D = p esc C. 

logp = 2.68124 

log esc C = 0.22613 



logD = 2.90737 
D= 807.9. 



Exercise V. Page 347. 

1. Given If 38° 14' N., L" 39° 51' N., X' 2° 7' E., X" 4° 18' E. ; required 
C and B. 

39° 51' N. Mer. parts = 2596.2 4° 18" E. 

38°14'N. =2471.8 2° 7" E. 



L d = 1°37' = 97' 
tan C ' 



Xd 



Mer. Ld 
logXd = 2.11727 

colog Mer. L d = 7.90518 
log tan C = 10.02245 

.-. C = N. 46° 29' E 



Mer. £<* = 124.4 \ d = 2° 11" = 131' 
B — Ld sec C 



logXd =1.98677 
log sec G = 0.16205 
log 2) =2.14882 
.\D = 140.9. 



teachers' edition. 333 

2. Given If 49° 53 7 N., L" 48° 28 7 N. , X 7 6° 19' W., X 77 5° 3 7 W. ; required 
C and D. 

49° 53' N. Mer. parts = 3446.0 6° 19 7 W. 

48° 28' X. Mer. parts = 3316.4 5° 3 7 W. 

L d - 1°25 7 = 85 7 Mer. L d = 129.6 Xd = 1° 16' = 76' 

tan C = ^ d _ • D = L d sec C. 

Mer. Ld 

\og\d =1.88081 logLa =1.92941 

colog Mer. L d = 7.88730 log sec C = 0.06416 

log tan C =9. 76829 log D = 1 . 99357 

.-. C=S. 30°23 / E. .-.2) = 98.5. 

3. Given X 7 64° 30' N. , L" 60° 40' N. , V 4° 20' W. , \" 0° W E. ; required 
C and D. 



64° 30 7 N. Mer. 


parts = 


5087.7 


4° 20' W. 


60° 40' N. Mer. 


parts = 


4582.2 


0° 10 7 E. 


X d = 3° 50' = 230' Mer. \ d - 


505.5 


\ d - 4° 30 7 = 270' 


f _ T1 r __ *<* 






Z> = Ld sec C. 


tan u — Ar _ 
Mer. Z d 


log X rf = 2.43136 






logLd =2.36173 


colog Mer. L d = 7.29628 






log sec C = 0.05569 


log tan C =9.72764 






logD =2.42742 


.-. C=S. 28° 24' 


E. 




.-. D = 261.5. 



4. Given Z, 7 54° 54' S., L" 34° 22 7 S., X' 60° 28 7 W., X 77 18° 24 7 W.; 
required C and D. 

54° 54' S. Mer. parts = 3938.7 60° 28' W. 

34° 22 7 S. Mer. parts = 2185.1 18° 24' W. 

L d = 20° 32' = 1232 7 Mer. L d = 1753.6 \ d = 42° 4' = 2524' 

tan C = ., X(Z - ■ D = L d sec C. 
Mer. L d 

\og\ d = 3.40209 logLd =3.09061 

colog Mer. L d = 6.75607 log sec C = 0.24376 

log tan C = 10.15816 log J) = 3.33437 

... C = X. 55° 13' E. .-. D = 2160. 

5. Given L' 17° 7 N., L" 20° 7 N. , X 7 180° 7 E. , X 77 177° 7 E. ; required 
C and D. 

20° 7 N. Mer. parts = 1217.3 180° 7 E. 

17° 7 N. Mer. parts = 1028.6 177° 7 E. 

L d - 3°0 7 =180 7 MerL d = 188.7 \ d = 30° 7 = 180 7 



334 NAVIGATION. 

tan C == X< * - B = L d secC. 
Mer. \d 

logXd =2.25527 \ogL d =2.25527 

colog Mer. L d = 7.72423 log sec C = 0.14052 

log tan C =9.97950 logD =2.39579 

C = N. 43° 39' W. D = 248.8, 

6. Given L' 45° 15' N., X' 35° 26' W., C N. 49° E., D 175; required 
Z"andX". 



La — B cos 0. 


\d — B sin C sec D m . 


logD =2.24301 


log J) = 2.24301 


log cos C = 9.81694 


log sin C = 9.87778 


logXd =2.05995 


log secD m = 10.15980 


D d = 115 / 


logXd = 2.27959 


= 1°55 / 


.-. \ d = 190' 


jj = 45° 15' 


= 3°ii' 


x „ - 470 10 / N 


X 7 = 35° 26' 


D m = 46°12'. 


X" = 32 15' W. 


7. Given X 7 55° l'N, Y 1° 25' 


E., ON. 10° E., D246; required U 


and X". 




L d = B cos C. 


55° V N., Mer. parts = 3950.9 


logD =2.39094 


59° 3' N., Mer. parts = 4395.3 


log cos C = 9.99313 


Mer. X^ = 444.4 


logD =2.38407 


\d — Mer. Ld x tan (7. 


B = 242' 


log Mer. i rf = 2.64777 


= 4°2 / 


log tan C = 9.24632 


£' = 55° r 


logXd =1.89409 


D" = 59°3 / N. - 


.-. \ d = 78 / 


L m = 57° 2'. 


= 1° 18' 



X 7 = 1° 25' 



X" = 2° 43' E. 

8. Given 2/ 50° 48' N. f X 7 9° 10' W., C S. 41° W., B 275 ; required L" 
and X". 

L d = B cos C. 50° 48' N., Mer. parts = 3532.0 

logD = 2.43993 47° 20' N., Mer. parts = 3215.2 

log cos C = 9.87778 Mer. L d = 316.8 

log L d = 2. 31771 Xd = Mer. L d tan 0. 

.-. L d = 208' log Mer. L d = 2. 50079 

= 3°28 / log tan C = 9.93916 

£' = 50° 48' log \ d = 2.43995 

£" = 47° 20' N. X d = 275' 

X w = 49°4 / . • = 4° 35' 

X' = 9°10 / 
X- = 13° 45' W. 



TEACHERS EDITION. 



335 







51° 18' N. 








37° O'N. 






U 


= 14° 18' 
= 858'. 




COS 


\C 


_L d 
D 




logL d 




= 2.93349 




logD 




= 3.01157 




log cos 


c 


= 9.92192 






c 


= N. 33° 20' 


w. 



9. Given L' 37° 0' N., L" 51° 18' N., V 48° 20' W., D 1027 ; required 
\" and C. 

Mer. parts = 3579.6 
Mer. parts = 2378.8 
Mer. L d = 1200.8 



\d = Mer. Ld tan C. 

log Mer. L d = 3.07947 
log tan C = 9.81804 
logX d =2.89751 

Xrf = 790' 
= 13° 10' 

V = 48° 20' 

X" = 61 o 30'W.or36 o 10'W. 

37° 5' N., V 9° 50' W., O S.W. by S.; 

Mer. parts = 3574.8 
Mer. parts = 2385.1 
Mer. L d =1189.7 

\d — Mer. Ld tan C. 
log Mer. L d - 3.07542 
log tan C = 9.82489 
logXd =2.90031 

Xd = 795' 

= 13° 15' 
y= 9°50 / 
\- - 23° 5' W. 



10. Given X 7 51° 15' N., 1/ 

required X" and D. 

51° 15' N. 
37° 5'N. 
X d = 14° 10' 

= 850'. 
C = 33° 45'. 
D = Ld sec C 
logX d =2.92942 
log sec C = 0.08015 
logD =3.00957 
D = 1022. 



11. Required the course and distance from Toulon to Valencia, by 
Mercator's sailing: 

= 39° 27' N. 
0° 19' W. 
43° 8'N. Mer. parts = 2858.3 5° 56' E. 

39° 27' N. Mer. parts = 2565.2 0° W W. 

L d = 3°41 / = 221 / Mer. L d - 293.1 



_ , fi = 43° 8'N. _. . fi = i 

Toulon (x= 5°56'E. Valencia ( x = 



3° 4r = 221' 

D — Ld sec C. 

logX rf =2.34439 
log sec C = 0.21050 
log D =2.55489 
D= 358.8. 



X rf = 6° 15' = 375' 

tanC= — ^— • 
Mer. X rf 

logXd = 2.57403 

log Mer. L d = 2.46702 

log tan C =10.10701 

C = S. 51° 59' W. 



336 



NAVIGATION. 



12. Kequired the compass course and distance from Cape East, New 
Zealand, to San Francisco; variation 14° 20' E., and deviation 5° 40' E.: 
\L = 37°40'S. _ _ fL = 37°48 / N. 

. X = 178° 36' E. San FranC1SC ° | X = 122° 24' W. 
37° 48' N. Mer. parts = 2439.0 178° 36' E. 

37° 40' S. Mer. parts = 2428.9 122° 24' W. 

:4528' 



Cape East < 



L d = 75° 28' : 



Mer. L d - 4867.9 



D — La sec G. 



\ d = 59° 0' = 3540' 

X d 



\ogL d =3.65591 
log sec C = 0.09222 
logD =3.74813 
D = 5599. 



tan C- , . 

Mer. La 

logXd =3.54900 

log Mer. L d = 3.68734 

log tan C =9.86166 

C = N. 36° 2' E. 

Var. and dev. = 20° 0' L. 

Compass course = N. 16° 2' E. 



Cape Lopatka i _ . 



13. Required the course and distance from Cape Lopatka to Callao : 
L= 51° 2'N. rX= 12° 4'S. 

156° 50' E. CaUa0 \\= 77°14'W. 

51° 2' N. Mer. parts = 3554.1 156° 50' E. 

12° 4' S. Mer. parts = 724.6 77° 14' W. 

L d = 63° 6' = 3786' Mer. L d = 4278. 7 234° 4' 

360° - 234° 4' = 125° 56' = 7556 m. = A* 



D — Ld sec C. 

\ogLd =3.57818 
log sec C = 0.30744 
logJD =3.88562 
D = 7685. 



tan C- 



X d 



Mer. Ld 

logXd = 3.87829 

colog Mer. L d = 6.36869 

log tan C = 10.24698 

O = S. 60° 29' E. 



14. A ship from latitude 20° 40' N. sails N.E. by N. until she is in 
latitude 27° 16' N. Required the distance and difference of longitude. 



27° 16' N. 




Mer. parts = 1691.0 


20° 40' N. 




Mer. parts = 1259.7 


L d = 6° 36' = 


396'. 


Mer. i d = 431.3 


C = 33° 45'. 






D — Ld sec C. 




\d — Mer. Ld tan C. 


\ogLd =2.59770 




log Mer. L d = 2.63478 


log sec C = 0.08015 




log tan G = 9.82489 


logD =2.67785 




logXd =2.45967 


D = 476.3. 




\ d = 288' = 4° 48'. 



TEACHERS EDITION. 



337 



15. A ship- from Cape Clear, in latitude 51° 26' N. and longitude 9° 29' 
W. , sails S. W. by S. until the distance run is 1022 miles. Find the lati- 
tude and longitude in by Mercator's and Middle Latitude Sailings. Which 
method is preferable ? 

By Mercator's Sailing, 

Mer. parts of 51° 26' = 3592.4 

Mer. parts of 37° 16' = 2398.8 







Mer. L d = 1193.6 




La — D cos C. 


\d = Mer. Ld tan C. 


D = 1022 


log D ~ 3.00945 


log Mer. L d = 3.07686 


C = 33° 45' 


log cos C = 9.91985 


log tan C =9.82489 




\ogLd =2.92930 


\og\d =2.90175 




L d = 850' 


\d = 798' 




= 14° 10' 


= 13° 18' 


♦ 


i' = 51° 26' 
2," = 37°"16' N. 


\'= 9° 29' 




X" = 22° 47' W. 




£ W =44°21'N. 


By Mid. Lat. Sailing, 

Xrf = D sin C sec L m 
logD =3.00945 
log sin C =9.74474 
log sec L m = 0.14564 


Mercator's 


> Sailing is preferable, 


\og\ d =2.89983 


since 


C<45°. 


\ d = 794' 
= 13° 14' 

X'= 9° 29' 




X- = 22° 43' W. 




Exercise VI. 


Page 351. 




1. 





a 


D. 


N. 


8. 


E. 


W. 


s.s.w. 

S.W. by S. 
N.E. 


2pts. 

3 pts. 

4 pts. 


48 
36 
24 


17 


44.3 
29.9 


17 


18.4 
20. 


Hence, L d = 57.2 S. 
= 0° 57' 


S. 

r 




74.2 
17. 


17 


38.4 

17. 


P 


= 21.4W 


57.2 


21.4 



338 



NAVIGATION. 









2. 




* 




a 


D. 


k 


8. 


K 


W. 


S.JE. 
S.W. is. 

s.s.w.jw. 


Jpt. 
3 i pts. 
2J pts. 


18 
37 
56 




17.9 

28.6 
50.6 


1.8 


23.5 
23.9 


Hence, L d = 97.1 S 
= 1°37' 


S. 
V. 





97.1 
0. 


1.8 


47.4 

1.8 


T 


-45.6\ 


97.1 


45.6 



a 


Z). 


N. 


s. 


E. 


W. 


S.S.W. i w. 
S.S.W. i w. 
S.byW. |W. 


2\ pts. 
2J pts. 
1J pts. 


43 
39 

27 




38.9 
34.4 

25.8 




18.4 
18.4 

7.8 


Hence, Z d = 99.1 S. 
- 1° 39' S. 







99.1 
0. 





44.6 
0. 


P = 


= 44.6 W. 


99.1 


44.6 



a 


D. 


N. 


S. 


E. 


W. 


N. 25° W. 
N. 8°E. 
N. 19° E. 
N. 76° E. 


16.4 

7.8 

13.7 

39.6 


14.9 

7.7 

13.0 

9.6 




0.1 
1. 

4.5 
38.4 


6.9 


Hence, L d = 45.2 N. 
= 0°45'N. 
p = 37.1E. 


45.2 

0. 





44. 
6.9 


6.9 


45.2 


37.1 



TEACHEKS' EDITION. 



339 




C. D. N. s. 


E. 


IT. 


S. 83° W. 
S. 48° E. 
N. 48° W. 
N. 36° W. 


23 
25.2 
27.1 
2.1 


18.1 
17. 


2.8 
16.9 


18.7 


22 8 

20.1 
12.3 


Hence, L d = 15.4 N. 
= 0° 15' N. 


35.1 
19.7 


19.7 


18.7 


55.2 
18.7 


p = 36.5 W. 




15.4 


36.5 



7. 



a 


D. 


K 


8. 


E. 


W. 


S. 17° E. 


48 




45.9 


14.0 




S. 45° W. 


19 




13.4 




13.4 


N. 36° W. 
N. 41° W. 


18 
50 


14.6 

37.7 






10.6 
32.8 


E. (90°). 


36 


.0 


.0 


36.0 




Hence, L d = 0° T S 




52.3 


59.3 
52.3 


50.0 


56.8 
50. 


p = 6.S W. 












7. 


6.8 



340 



NAVIGATION. 
8. 



a 


D. 


jsr: 


S. 


R 


W. 


N.N.E. 


2 pts. 


31 


28.6 




11.9 




E.N.E. 


6pts. 


35 


13;4 




32.3 




E. by S. 


7 pts. 


36 




7 


35.3 




S.S.E. 


2 pts. 


51 




47.1 


19.5 




S. by E. 


lpt. 


60 




58.8 


11.7 




Hence, L d = 70.9 


- 


42.0 


112.9 


110.7 




= 1°11' 


S. 
E. 




42.0 






T 


= 110.7 


70.9 



9. 



o. 


D. 


N. 


S. 


& 


W. 


S. 44° E. 
S. 85° E. 
S. 27° E. 
N. 37° W. 
N. 20° W. 


69 
68 
25 
5 
13 


4.0 
12.2 


49.6 

5.9 

22.3 


47.9 
67.7 
11.3 


3.0 
4.4 


Hence, L d = 61.6 

= 1° 2' S. 


16.2 


77.8 
16.2 


126.9 
7.4 


7.4 


p = H9.5 : 


E. 


61.6 


119.5 



Exercise VII. Page 359. 

1. First course : N.N.E. = 2 points R. of N. = N. 22° 30 / E., 31.4 m. 

Second course : E.N.E. = 6 points R. of N. = N. 67° 30 / E., 35 m. 

Third course : E. by S. = 7 points L. of S. = S. 78° 45 / E., 36.1 m. 

Fourth course : S.S.E. = 2 points L. of S. = S. 22° 30 / E., 50.9 m. 
Tide course : = 1 point L. of S. = S. 11° 15/ E., 60 m. 



TEACHERS EDITION. 



341 



The Traverse. 



a 


D. 


jsr. 


S. 


R 


W. 


N.N.E. 


2pts. 


31.4 


29. 




12. 




E.N.E. 


6 pts. 


35. 


13.4 




32.3 




E.by S. 


7pts. 


36.1 




7. 


35.4 




S.S.E. 


2 pts. 


50.9 




47. 


19.5 




S. by E. 


1 pt. 


60. 




58.8 


11.7 




L d = 70.4' 


= PIO'S. 


42.4 


112.8 


110.9 


= P- 


If 

L" 


= 46° 28' N. 




42.4 






= 45° 18' 


N. 


70.4 



\ d - 159.3' 
= 2° 39' E. 

X' = 22° 18' W. 
X" = 19° 39' W. 

= 11° 15' R. of S. 
12° 20' L. of S. 



1° 5' L. of S. 
, 40 m. 



p — 110.9 Kt = p sec L m . 

L m - 45° 53' logp = 2.04493 

log sec L m = 0.15731 
logX d = 2.20224 

2. First course: 

S. by W.= lpt. R. of 8 

Variation 

True course 

Hence, course and distance S. 1° 5' E 
Second course (starboard tack) • 

S.W. by S.= 3pts. R. of S. 
Leeway = 1 pt. L. 

2 pts. R. of S. = 22° 30' R, of S. 

Variation 12° 20' L. 

True course 10° 10' R, of S. 

Hence, course and distance S. 10° 10' W., 69.6 m. 
Third course : 

S.W. by W.= 5 pts. R. of S.= 56° 15' R. of S. 

Variation 12° 20' L. 

True course 43° 55' R. of S. 

Hence, course and distance S. 43° 55' W., 58.5 m. 

Current course: 

W.S.W.= 6 pts. R. of S.= 67° 30' R. of S. 

Variation 12° 20' L. 

True course 55° 10' R. of S. 

Hence, course and distance S. 55° 10' W., 36 m. 



342 



NAVIGATION. 



The Teaveese. 



a 


D. 


K 


8. 


E. 


W. 


S. 1°E. 




40. 




40. 


0.7 




S. 10° W. 




69.6 




68.6 




12.1 


S. 44° W. 




58.5 




42.1 




40.7 


S. 55° W. 




36. 




20.6 




29.5 


id = 171.3' 
IS 
L" 


= 2°51'S. 
= 33° 40' N. 




171.3 


0.7 
P = 


82.3 
0.7 


= 30° 49 


'N. 


81.6 



p = 81.6 

L m = 32° 15' 



\d = P sec L m 
\ogp= 1.91169 
log sec L m — 0.07277 
\og\ d = 1.98446 



\ d = 96' 

= 1°36'W. 
y = 16° 20' w. 



X" = 17°56 / W. 



3. First course (starboard tack) : 

N. by E.= lpt. K. of N. 

Leeway = 1 pt. L. 

= due north = 0° 
Variation . . . . W. 13° 30' L. 



13° 30' L. of N. 
Hence, course and distance N. 13° 30' W., 37.7 in. 

Second course (starboard tack) : 

N.= 0°pt. 
Leeway = 1 pt. L. 

lpt. L.= li°15'L. of N. 
Variation . . . . W. 13° 30' L. 

24° 45' L. of N. 
Hence, course and distance N. 24° 45' W., 38.7 m. 

Third course (starboard tack) : 

N.N.W.= 2pts. L. of N. 

Leeway = 1 pt. L. 

3 pts. L. of N. = 33° 45' L. of N. 
Variation . , W. 13° 30' L. 

47° 15' L. of N. 
Hence, course and distance N. 47° 15' W., 76.5 m. 



TEACHERS' EDITION. 



343 



Current course : 

W.N. W. =6 pts. L. of N. = 67° 30' L. of N. 
Variation 13° 30' L. 



81° <y L- of N. 
Hence, course and distance N. 81° "W., 12 m. 

The Traverse. 



a 


D. 


#". 


S. 


E. 


W. 


N. 14° W. 


37.7 


36.6 






9.2 


N. 25° W. 


38.7 


35. 






16.4 


N. 47° W. 


76.5 


52.1 






56. 


N. 81° W. 


12. 


1.9 






11.9 


i d =125.6 / = 2° 6'N. 


125.6 




P = 


93.5 


JJ = 19° 30' S. 










L" - 17° 24' S. 











p = 93.5 

L m = 18° 27' 



\d = i) sec i w . 

logp =1.97081 

log sec L m = 0.02296 
logXd =1.99377 



X d = 99' 

= 1° 39' W. 

y = QQ 1Q- e. 
X // _ p 29^ w. 



4. Departure course (the opposite of W.S.W.): 

E.N.E. The ship's head S.E. by E.; the deviation is the same as for 
the first course. 

E.N.E. = 6 pts. R. of N.= 67° 30' R. of N. 

Variation and deviation . . 17° L. 

50° 30' R. of N. 
Hence, course and distance N. 50° 30' E., 18 m. 

First course: 

S.E. by E.= 5 pts. L. of S.= o6° 15' L. of S. 
Variation and deviation . . . 17° L. 

True course 73° 15' L. of S. 

Hence, course and distance S. 73° 15' E., 52 m. 
Second course (port tack) : 

S.E. = 4 pts. L. of S. 

Leeway = j pt. R. 

3* pts. L. of S.= 39° 22' L. of S. 
Variation and deviation . . . 19° L. 
True course 



58° 22' L. of S. 
Hence, course and distance S. 58° 22' E., 43 m. 



344 NAVIGATION. 

Third course (starboard tack) : 

E. by N. = 7 pts. R. of N. 
Leeway = 1 pt. L. 



6 pts. R. of N. = 67° 30' R. of N. 
Variation and deviation . . 11° L. 



True course . . . . . 56° 30" R. of N. 
Hence, course and distance N. 56° 30' E., 36 m. 

Fourth course (starboard tack) : 

E.N.E. -6 pts. R. of N. 
Leeway = 1\ pts. L. 

4i pts. R. of N.= 50° 37' R. of N. 
Variation and deviation . . 13° L. 



True course . . . . . 37° 37' R. of N. 
Hence, course and distance N. 37° 37' E., 27 m. 

Fifth course (port tack) : 

S.S.E.= 2pts. L. of S. 
Leeway = 2 pts. R. 



pts. = due south = 0° 
Variation and deviation . . 21° L. 



True course 21° L. of S. 

Hence, course and distance S. 21° E., 24 m. 

Sixth course (port tack) : 

S.E. by S.= 3 pts. L. of S. 
Leeway = 1£ pts. R. 

If pts. L. of S. = 19° 41' L. of S. 
Variation and deviation . . . 20° L. 



True course 39° 41' L. of S. 

Hence, course and distance S. 39° 41/ E., 29 m. 

Current course : 

S. by E.= 1 pt.= L. of S.= 11° W I" of s - 
Variation 28° L. 



39° 15' L. of S. 
Hence, course and distance S. 39° 15' E,, 12 m. 



TEACHERS' EDITION. 



345 



The Traverse. 



a 


D. 


N. 


S. 


E. 


W. 


N. 51° E. 


18 


11.3 




14.0 




S. 73° E. 


52 




15.2 


49.7 




S. 58° E. 


43 




22.8 


36.5 




N. 57° E. 


36 


19.6 




30.2 




N. 38° E. 


27 


21.3 




16.6 




S. 21° E. 


24 




22.4 


8.6 




S. 40° E. 


29 




22.2 


18.6 




S. 39° E. 


12 




9.3 


7.6 




L d = 40' S. 

JJ - 47° 31' N. 


52.2 


91.9 
52.2 


181.8 


-P. 


£" = 46° 5r 


N. 


39.7 



p = 181.8 
L m = 47° 11' 



X rf = p sec L m 
logp =2.25959 

logsecZ m = 0.16771 



X d = 267' 

= 4°27E. 
V = 52° 33' W. 



logXd 



= 2.42730 



\ // = 48° 6 7 W. 



5. Departure course (the opposite of W. by S. i S.): 

E. by N. i N. = 6f pts. R. of N. 
= 75° 56' R. of N. 
Variation and deviation . . 34° R. 



109° 56' R. of N. 
Hence, course and distance S. 70° E., 17 m. 

First course (port tack) : 

S.S.E.= 2 pts. L. of S. 
Leeway = 2£ pts. R. 



ipt. R. of S.= 2°49 / R. of S. 
Variation and deviation . . . 34° R. 



36° 49 7 R. of S. 
Hence, course and distance S. 37° W., 21 m. 



346 NAVIGATION. 

Second course (starboard tack) : 

S.S. W. i W. = 2i pts. R. of S. 
Leeway = 2f pts. L. 



ipt. L. of S.= 2°49'L. of S. 
Variation and deviation . . . 27° R. 

24° 11' R. of S. 
Hence, course and distance S. 24° W. , 20 m. 

Third course (port tack) : 

W.S.W.= 6 pts. R. of S. 
Leeway = 2j pts. R. 

8* pts. R. of S.= 7i pts. L. of N. 
= 84° 22' L. of N. 
Variation and deviation . . . 22° R. 



True course 62° 22' L. of N. 

Hence, course and distance N. 62° W. , 24 m. 

Fourth course (starboard tack) : 

W. i N. = 7£ pts. L. of N. - 84° 22' L. of N. 
Variation and deviation . . . 20° R. 



True course 64° 22' L. of N. 

Hence, course and distance N. 64° W. , 26 m. 

Fifth course (starboard tack) : 

East = 8 pts. R. of N. 

Leeway = 2j pts. L. 

6* pts. R. of N. = 61° 52' R. of N. 

Variation and deviation . . 41° R. 

True course 102° 52' R. of N. 

Hence, course and distance S. 77° E., 19 m. 

Sixth course (starboard tack) : 

E.S.E.= 6 pts. L. of S.= 67° 30 r L. of S. 
Variation and deviation . . . 40° R. 



True course 27° 30' L. of S. 

Hence, course and distance S. 28° E., 18 m. 

Current course: N.N.E.= 2 pts. R. of N. 

= 22° 30' R. of N. 
Variation 31° R. 



True course 53° 3CT R. of N. 

Hence, course and distance N. 54° E., 21 m. 



TEACHEKS EDITION. 



347 



The Traverse. 



a 


D. 


N. 


& 


E. 


W. 


S. 70° E. 


17 




5.8 


16.0 




S. 37° W. 


21 




16.8 




12.6 


S. 24° W. 


20 




18.3 




8.1 


N. 62° W. 


24 


11.3 






21.2 


N. 64° W. 


26 


11.4 






23.4 


S. 77° E. 


19 




4.3 


18.5 




S. 28° E. 


18 




15.9 


8.5 




N. 54° E. 


21 


12.3 




17. 




L d = 26' S. 
1/ =62° O'N. 


35.0 


61.1 
35.0 


60. 


65.3 
60. 


IT = 61° 34' 


N. 


26.1 


5.3 



p =5.3 . 

L m = 61° 47' 



\d — p sec L m . \a — ll 7 W. 

logp = 0.72428 V = 150° 7 E. 

log sec L m = 0.32532 \" = 149° 49 7 E. 
log\ d =: 1.04960 



6. Departure course (the opposite of N. £ W.): 

S. f E. = f of a pt. = 8° 26 r L. of S. 
Variation and deviation . . . 8^_ R. 

0° 26 7 L. of S. 
Hence, course and distance S., 19 m. 
First course (port tack) : 

S.W. i W.= H pts. R. of S. = 50° 37 7 R. of S. 
Variation and deviation . . . 8° R. 

True course 58° 37 7 R. of S. 

Hence, course and distance S. 59° W. , 58 m. 
Second course (starboard tack) : 

N. f E. = f pt. R. of N. 

Leeway = Sj pts. L. 

2i pts. L. of N. = 28° V L. of N. 

Variation and deviation . . . 17° R. 

True course 11° V L. of N. 

Hence, course and distance N. 11° W., 15 m. 



348 NAVIGATION. 

Third course (starboard tack) : 

S.E. £E.= lipts. L. of S. 
Leeway — 2£ pts. L. 



4± pts. L. of S,= 47° 48' L. of S. 
Variation and deviation . . . 20° R. 



True course . . . . . 27° 48' L. of S. 
Hence, course and distance S. 28° E., 9 m. 

Fourth course (port tack) : 

W. by S.= 7 pts. R. of S. 
Leeway = i pt. R. 

7± pts. R. of S. = 81° 33' R. of S. 
Variation and deviation ... 0° 



True course 81° 33" R. of S. 

Hence, course and distance S. 82° W., 50 m. 

Fifth course (starboard tack) : 

E.N.E.= 6 pts. R. of N. 
Leeway = 2-J pts. L. 

3i pts. R. of N.= 39° 22' R. of N. 
Variation and deviation . . . 33° R. 



True course 72° 22' R. of N. 

Hence, course and distance N. 72° E., 12 m. 

Sixth course (port tack) : 

S.S.W. i W.= 2i pts. R. of S. 
Leeway = If pts. R. 



4 1 pts. R. of S.= 47° 48 r R. of S. 
Variation and deviation . . . 10° R. 



True course 57° 48' R. of S. 

Hence, course and distance S. 58° W., 22 m. 

Current course : 

S.W. i W.= 4 J pts. R. of S.= 47° 48' R. of S. 
Variation 14° R. 



True course 61° 48 7 R. of S. 

Hence, course and distance S. 62° W., 42 m. 



TEACHERS EDITION. 



349 



The Traverse. 



p= 142.3 
L m = 50° 51' 



\d — p sec Lm 

logp =2.15320 

logsecZ m = 0.19973 



c. 


I). 


N.. 


A 


E. 


W. 


s. 




19 




19.0 






S. 59~ TV 




58 




29.9 




49.7 


N. 11° W. 




15 


14.7 






2.9 


S. 28° E. 




9 




7.9 


4.2 




S. 82° W. 




50 




7.0 




49.5 


N. 72° E. 




12 


3.7 




11.4 




S. 58° W. 




22 




11.7 




18.7 


S. 02° W. 




42 




19.7 




37.1 


Ld — 77' = 
1/ = 


1° 17' S. 

50° 12' S. 


18.4 


95.2 
18.4 


15.6 


157.9 
15.6 


51° 


29' S. 


76.8 


142.3 



\d = 225' 

= 3° 45' W. 
y = 179° 40' W. 



logXj 






5293 



X" = 176° 35' E. 



7. First course : 

S.E.= 4 pts. L. of S. 
Variation and deviation . . 1| pts. L. 



True course . 5J pts. L. of S. 

Hence, course and distance S. 5-J-^pts. E., 27.8 m. 

Second course : 

E.S.E. JE.= 6Jpts. L. of S. 
Variation .... If pts. L. 

True course .... 8 pts. L. of S. = due east. 
Hence, course and distance E. 75.2 ni. 

Third course : 

E.= 8 pts. R. of N. 
Variation and deviation . . 1£ pts. L. 

True course .... 



6£ pts. R. of N. 
Hence, course and distance X. 6f pts. E., 8.7 m. 



350 



NAVIGATION. 



The Traverse. 



c. 


D. 


N. 


8. 


E. 


W. 


S. 5£ pts. E. 

E. 

N. 6f pts. E. 


27.8 
75.2 

8.7 


2.1 


13.1 


24.5 
75.2 

8.4 




L d = ir s. 

U = 36° 42' N. 


2.1 


13.1 
2.1 


108.1 


= P- 


2/' =36° 31' 


N. 


11.0 



p = 108.1 
L m = 36° sr 



\ d = p sec L m 
logp =2.03383 

log sec L m = 0.09548 
logXd =2.12931 



\ d = 135' 

= 2° 15' E. 
V = 4° 25' W. 



Exercise VIII. Page 374. 

1. Find the elements (initial courses, distance, and latitude and longi- 
tude of the vertex) of the great circle track between the Lizard, in latitude 
49° 58' N., longitude 5° 12' W., and the Bermuda Islands, in latitude 32° 
18' N., longitude 64° 50' W. 

Referring to the triangle CPC, 

Lat. C" = 49° 58' N. 

Long. C" = 5° 12' W. 

Lat. C = 32° 18' N. 

Long. C = 64° 50' W. 
c = 90° - 49° 58' = 40° 2'. 
c' = 90° - 32° 18' = 57° 42'. 
\ d = 64° 50' - 5° 12' = 59° 38'. 




To find the initial courses : 

tan£(C"+ C) 

tan£(C- C) 



— cos j (c' — c) 
cos i (c' + c) 

— sin ^ (c' — c) 
~ sin i (c' -f- c) 



cot i \ d - 
cot i Xd. 



teachers' edition. 



351 



±(c'-c)= 8° 50', 
$ ( c ' + c) = 48° 52', 

}Xd =29° 48', 



log cos = 9.99482 
cologcos = 0.18190 
log cot = 10.24178 



log tan i (C + C) = 10.41850 
i(C' + G') = 69° 7' 
$(C- C) = 19° 35' 
... (?' = N. 88° 42' W 
C = X. 49° 32' E. 
To find the distance : 

cos iD~ 

i (c + c') = 48° 52', 
l(C+ C') = 69° 7', 
i\ d =29° 49', 

iD = 23°26'. 
D = 46° 52' : 
To find L of F 
sin TV = sin c' sin C. 
log sine' =9.92699 
log sin C = 9.88126 
log sin PV= 9.80825 
PF = 40°r. 
X of V = 90° - 40° 1' 
= 49° 59' N. 



log sin = 9.18628 

cologsin= 0.12310 

log cot = 10.2417 8 

log tan i(C'-C) = 9.55116 

i(C- C) = 19° 36'. 

= course from Lizard. 
= course from Bermudas. 



cos \(c + c') . f ^ 
coBi(C+C') 8miXj - 




log COS 


= 9.81810 


colog COS 


= 0.44798 


log sin 


= 9.69655 



logcos£Z> = 9.96263 

2812 m. 

To find X of F 
cot CPV = cos c' tan C. 
log cose' = 9.72783 
log tan C = 10.06901 
log cot CPF= 9.79684 
CPF = 57° 56'. 
X of C = 64° 50' 
57° 56' 
X of F = 6° 54' W. 



2. Find the elements of the great circle track between Boston (Minot's 
Ledge light-house) in latitude 42° 16' N., longitude 70° 46' W., and Cape 
Clear, in latitude 51° 26' N., longitude 9° 29' W. [Take i \ d = 30° 39'.] 





Lat. C 


= 51° 


26' N. 










Long. C" 


= 9° 


29' W. 










Lat. C 


= 42° 


16' N. 










Long. C 


= 70° 


46' W. 








c 


= 90° - 1 


31° 26' 


= 38° 


34'. 






€ 


= 90° - 42° 16' 


= 47° 


44'. 






\d 


= 70° 46' 


-9°: 


26' = 61° 1' 


V. 


c 




tan 


i(C- 


+ C) = 


cos 

COS 




-c) 
-he) 




tan 


f(C 


-C) = 


sin 
sin 




-c) 
+ c) 




cot -J- Xd. 

cot -J- X<*. 



352 



NAVIGATION, 



£( C '- C )= 4°35', log cos = 9.99861 

f (c?-rhe) = 43° 9', cologcos = 0.13694 

£X d = 30° 39', log cot = 10.22726 

log tan i (C + C) = 10.36281 

^ (C + 0) = 66° 33' 

i(C — C) = 11° 9 / 



log sin = 8.90260 

cologsin= 0.16500 

log cot = 10.22726 

log tan i (C - C) = 9.29486 

£ (C' - G) = 11° 9'. 



,.. C' - N. 77° 42' W.= course from Cape Clear. 
C — N. 55° 24' E. = course from Boston. 



To find the distance : 



cos i B = 



cos j (c' ,+ c) 



sin i \ d . 



cos±(C" + (7) 
^ ( C ' + c) = 43° 9', log cos 

I (C .+ C) = 66° 33', colog cos 

} \ d = 30° 39', log sin 



= 9.86306 
= 0.40017 
= 9.70739 



...£2) =30° 50'. 
B = 41° 40 / 

To find L of V. 
sin PV — sin c' sin C 
log sin c' =9.86924 
log sin C = 9.91547 
log sin PV- 9.78471 
... PF = 37°32 / . 
X of V = 90° - 37° 32' 
= 52°28'N. 



log cos £D= 9.97062 

2500 m. 

To find X of V. 
cot CPV = cose' tan C. 

log cose 7 = 9.82775 
log tan C = 10.16124 
log cot CPV = 9.98899 
CPF= 45° 44' 
Xof C = 70° 46' 
Xof F=25° 2' W. 



3. Find the elements of the great circle track between Vancouver 
Island, in latitude 50° N., longitude 128° W., and Honolulu, in latitude 
21° 18' N., longitude 157° 52' W. 

Lat. C" = 50° N. 
Long. C = 128° W. 
Lat. C - 21° 18 / N. 
Long. C = 157° 52' W. 

c = 90° - 50° = 40°. 
V c' = 90° - 21° 18' = 68° 42'. 




\ d = 157° 52' - 128° = 29° 52'. 



tan *((?' + C) = 



cos j (c' — c) 
cos i (c' + c) 



cot I X<*. 



tan i (C - O) = C0S |^ , c j cot iX d . 
' v ■ sin i (c' + c) 



teachers' edition. 



353 



i ( C ' - c) = 14° 21', log cos = 9.98623 

J(c / + c) = 64°2r, colog cos = 0.23446 

iXd = 14° 56', log cot = 10.57397 

log tan i (C + G) = 10.79466 

*(C" + C) = 80° 53' 
■fr((7-C)= 48°50 / 
C" = 129° 43' 
= S. 50° 17' W. = course from Vancouver. 



log sin = 9.39418 

colog sin = 0.09013 

log tan = 10.57397 

log tan i (C - C) = 10.05828 

i(C"-C) = 48°50'. 



C = N. 32 
To find the distance : 

cos -J- D = 



3' E. = course from Honolulu. 



cos j (c' + c) 



sin \ X d . 



i (c' + c) = 54° 21', 
i (C + O) = 80° 53', 

*X<* = 14° 56', 



cosi(C / + C) 

log cos 



£D=18°34'. 
D = 37° $'■• 
To find X of V. 
sin PF • = sin c' sin C 
log sine 7 =9.96927 
log sin C = 9.72482 
log sinPV = 9.69409 
PV = 29° 38'. 
X of F = 90° — 29° 38' 
= 60° 22 / N. 



colog COS 
log sin 
log cos i D 



= 9.76554 
= 0.80012 
= 9.41110 
= 9.97676 



2228 m. 

To find X of V. 
cot CPV = cos € tan C. 
log cose' =9.56020 
log tan C = 9.79663 
log cot CPV- 9.35683 
CPV= 77° 11' 
X of C = 157° 52' 
X of V = 80° 41' W. 



4. Find the elements of the great circle track between Cape Clear, in 
latitude 51° 26' N., longitude 9° 29' W., and Sandy Hook, in latitude 
40° N., longitude 74° W. 

Lat. C" = 51° 26' N. 
Long. C" = 9° 29' W. 
Lat. C = 40° N. 
Long. C = 74° W. 

c = 90° - 51° 26' = 38° 34'. 

c' = 90° - 40° = 50°. 

\ d = 74° - 9° 29' = 64° 31'. 




i ,n » ™ cos ^(c—c 7 ) . T _ 

tan i(C+ C") = ;) , ' cot i X* 

v ' cos -J- (c + c') 



tan i (C - C") = 



sin j (c — c') 
sin i (c + c') 



cot i Xd . 



354 



NAVIGATION. 



K C ' — C )= 5043^ log cos = 9.99784 
$ (c + C) = 44° 17', colog cos = 0.14515 
i \ d = 32° 16', log cot = 10.19972 



log sin = 8.99830 
colog sin = 0.15602 
log cot = 10.1997 2 
log tan $(.C'—C)= 9.35404 
i(C"- C) = 12°44'. 



log tan i(C+ C") = 10.34271 
j (C + 6 f/ ) = 65° 34' 
i (C - C) = 12° 44' 

C = N. 78° 18' W. = course from Cape Clear. 
C = N. 52° 50' E. = course from Sandy Hook. 

To find the distance : 

1 cos j (c + c') . n x 



l(c + c') =44° 17', 
i(C+ C") = 65°34 / , 
i\ d = 32° 16', 



log cos = 9.85485 
colog cos = 0.38338 
log sin = 9.72743 
log cos iD= 9.96566 

£D = 22°29 / . 
B = 44° 58' = 2698 m. 



To find L of V. 
sin PV = sin c sin C". 
log sine =9.79478 
log sin C" = 9.99088 
log sin PF= 9.78566 
PF=37°37'. 
£ of F = 90° 00'- 37° 37' 
= 52° 23' N. 



To find X of V. 
cot CPV = cos c' tan C. 
log cose' = 9.80807 
log tan (7 = 10.12026 
log cot CPV = 9.92833 
CPF= 49° 42' 
X of C = 74° 0' 
Xof F=24°18' W. 



5. Find the elements of the great circle track between Lizard Light, 
in latitude 49° 58' N., longitude 5° 12' W., and Cape Frio, in latitude 
23° S., longitude 42° W. 



p 




Lat. C = 49° 58' N. 
Long. C" = 5° 12' W. 
Lat. C = 23° S. 


V ' 




Long. C = 42° W. 


jL~» V 




c = 90° - 49° 58' = 40° 2'. 


-^c' 




C ' = 90° + 23° = 113°. 

\ d = 42° - 5° 12' = 36° 48'. 


tan*(C' + 0) = 


_ cos 

COS 


i (C + c) 


tan \ (C - C) - 


_ sin 
sin 


tt + t cotn *- 



TEACHERS EDITION. 



355 



i ( C ' - c) = 36° 29', 
£ (c' + c) = 76° 31', 
^X rf =18° 24', 



log cos = 9.90o27 
colog cos = 0.63234 
log cot = 10.47801 



logtani(C + C) = 11.01562 
4(0'+ C) = 84°29' 
i(G"-C) = 61 27' 



log sin = 9.77421 

colog sin = 0.01214 

log cot = 10.4780 1 

log cot i (C - C) = 10.26436 

i(C— C) = 6P27 / . 



C'=S. 34^ 
C = X. 23° 



4' W. = course from Lizard Light. 
2' E. = course from Cape Frio. 



To find the distance : 



1 co s | (c' + c) . 
cos | Z> = . ' , r . sin i X<|. 

cos i(C -f C) 



j ( C ' + c) = 76° 31', 
| (J + c) = 84° 29', 
•}X rf = 18° 24', 



log cos = 9.36766 
colog cos = 1.01712 
log sin = 9.49920 
log cos £2)= 9.88398 
| 1) - 40° 3'. 
i> = 80° 6' = 4806 m. 



To find L of V. 
sin P T T = sin c' sin C. 
sin 180° - 113° = sin 67° = c\ 
log sin & =9.96403 
log sin C = 9.59247 
log sin PV= 9.55650 
Pr= 21° 7'N. 
90° 00' N. 
21° VX. 
L of V = 68° 53' N. 



To find X of T. 
cot CPV = cos c' tan C. 
= 9.59188(h) 
= 9.62855 



log cose" 
log tan C 
losr cot CPT 



= 9.22043(h) 
CPr= 99° 26' 
X of C = 42° 00' 
X of V= 57° 26' E. 



6. Find the elements of the great circle track between Cape Frio and 
Cape Good Hope, in latitude 34° 20' S., longitude 18° 30' E. (Reckon 
from the nearest pole.) 



Lat. C" = 34° 20' S. 

Long. C = 18° 30 / E. 

Lat. C = 23° S. 

Long. C = 42° W. 
c = 90° - 43° 20' = 55° 40'. 
c' = 90° - 23° = 67°. 
\ d = 42° + 18° 30' = 60° 30'. 



V C 




tan i (C + C) = C0 **V c \ cot | \ d . 
v ' cos i (c + c) 

teni(C - C) = S | n t| C !~ C ! cot iX d . 
v ' sm i (c' 4- c) 



356 



NAVIGATION. 



i (c' — c) = 5° 40", log cos 
^ ( c ' + c) = 61° 20', colog cos 
^Xd = 30° 15', log cot 

log tan i (C + C) 



9.99787 

0.31902 

10.23420 

10.55109 

| (C + C) = 74° 18' 
£(<7- 0) = 10° 55 7 



log sin = 8.99450 

colog sin = 0.05679 

log cot = 10.23420 

log tan (C— C) = 9.28549 



C - S. 85° 13' W.= course from C. Good Hope. 
C = S. 63° 23' E. = course from Cape Frio. 



To find the distance : 



i t,_ coslfc' + c) 



i (c' + c) = 61° 20', log cos 

i (& + C) = 74° 18', colog cos 

i \ d - 30° 15', log sin 

log cos £D= 9.95089 
iJD=26°44 / . 
I) - 53° 28' = 3208 m. 



= 9.68098 
= 0.56767 
= 9.70224 



To find L of V. 
sin PV = sin c' sin C. 
log sine' =9.96403 
log sin C = 9.95135 
log sin PV = 9.91538 
PF=55°23' 
XofF=90° 0'-55°23' 
= 34° 37' S. 



To find X of V. 
cot CPV - cos of tan C. 
log cose 7 = 9.59188 
log tan O = 10.30005 
log cot CPV= 9.89193 
CPF= 52° 3' 
42° 0' 
Xof F=10°3 / E. 



7. Find the elements of the great circle track between Grand Port, 
Mauritius, in latitude 20° 24' S., longitude 57° 47' E., and Perth, in lati- 
tude 32° 3' S., longitude 115° 45 / E. 

Lat. C" = 32° 3' S. 
Long. C - 115° 45' E. 
Lat. = 20° 24' S. 
Long. C = 57° 47 / E. 

c =90° -32° 3' =57° 57'. 
c' = 90° - 20° 24' = 69° 36'. 
\ d = 115° 45' - 57° 47' = 57° 58'. 




tan i (C + C) = 
tan£(C"- C)- 



cos j (c' — c) 
cos i (c' + c) 
sin j (c' — c) 
sin i (c' + c) 



cot i \ d . 



cot i \ d . 



teachers' edition. 357 

I ( C ' — C ) = 5° 50 , log cos = 9.99776 log sin = 9.00704 

| ( C ' + c) = 63° 47', colog cos = 0.35481 colog sin = 0.04714 

£ \ d = 28° 59', log cot = 10 .25655 log cot = 10.2565 5 

log tan i (C + C) = 10.60912 log tan £ (C - C) = 9.31078 

£ (C + C) = 76° 11' i (C - C) = 11° 34'. 

t(C- C) = ll°34 / 

(7 = s. 87° 45" E. = course from Perth. 
C = S. 64° 37' E. = course from Mauritius. 

To find the distance : 

cos j (c' + c ) 



(JUS -% u — 


COS £(G + C) 




£ {cf + c) = 63° 47', 


log COS 


= 9.64579 


£(C"+ C) = I6°ir, 


colog COS 


= 0.62194 


i\ d =28° 59', 


log sin 


= 9.68534 



log cos £2) =9.95247 
£D=26° 19'. 
D = 52° 38' = 3158 m. 

To find L of F. To find X of F 

sin PV — sin & sin C. cot CPV = cos c' tan C 

log sine' =9.97187 log cose' = 9.54229 

log sin C - 9.95591 log tan C = 10.32378 

log sinPF = 9.92778 log cot CPV = 9.86607 
PV = 57° 52'. CPF = 53° 42' 

X of F = 90° 0' - 57° 52' X of C = 57° 47' 

= 32° 8' S. X of F = 111° 29' E. 



8. Eind the elements of the great circle track between A, in latitude 
16° 38' N., longitude 70° 55' W., 
and B, in latitude 48° 2' N., lon- 
gitude 4° 35' W. 

LatC" =48° 2'N. 
Long. C - 4° 35' W. 
Lat. C = 16° 38' N. 
Long. C = 70° 55' W. 

c =90° -48° 2' = 41 a 58'. 
c' = 90° - 16° 38' = 73° 22'. 
\ d = 70° 55' - 4° 35' = 66° 20'. 




358 



NAVIGATION. 



tan i\C + C) = 



tan % (C — C) 



cos -j- ( c' — c ) 
cos -J- (c' + c) 
— sin j (c' — c) 



cot^Xd. 
cot i \ d . 



sin i (e' H- c) 

|(c' - c) = 15° 42', log cos = 9.98349 log sin : 9.43233 

i (& + c) = 57° 40', colog cos = 0.27177 colog sin = 0.07317 

$ \ d = 33° 10', log cot = 10.18472 log cot = 10.18472 

log tan i (C +C) = 10.43998 log tan i (C - C) = 9.69022 

i (C 4- C) = 70° 3' %\C'-ti) = 26° 6'. 

j(C / -C) = 26° 6 7 

C" = K ! 



9' W. 

= S. 83° 51' W.= course from B. 
C = N. 43° 57' E. = course from A. 



To find the distance : 



cos i B = 

£ ( C ' + c) = 57° 40% 
i(C , + C) = 70° 3', 
i\d = 33° 10', 



_ cos j (c' + c) 



cos •} (c' — c) 



£D=30°57'. 

j) = 61° 54' = 3714 m. 



sin i \ d . 

log cos =9.72823 
colog cos = 0.46699 
log sin = 9.73805 
log cos£D= 9.93327 



To find L of V. 
sin PV = sin c' sin C. 
log sine =9.98144 
log sin C" = 9.84138 
log sin PV- 9.82282 
PF=41°41'. 
ZofF=90° / -41 c 
= 48° 19' N. 



4r 



To find \ d of V. 
cot CPV = cos c' tan C. 
log cose' =9.45674 
log tan C = 9.98408 
log cot CP V= 9.44082 
CPV= 74° 34' 
X of C = 70° 55' 
X fF= 3°39 / E. 



9. A ship sails from A., in latitude 40° S., longitude 148° 30' E., to B, 

C in latitude 12° 4' S. , longitude 77° 14' W. 
Compare the great circle and the rhumb- 
line between A and B. 

Lat. C" - 40° S. 
Long. C" = 148° 30' E. 
Lat. C = 12° 4' S. 
Long. C = 77° 14' W. 

c = 90° -40° =50°. 
c' = 90° - 12° 4' = 77° 56'. 
\ d = 148° 30' + 77° 14' 
= 225° 44', or 134° 16'. 




TEACHERS EDITION. 



359 



tan i (C + C) = C08 *S^ , * j cot i X d . 
v ' cos £ (c' + c) 8 



tani(C / - C) = 



sin \ (c' — c) 



| ( C ' - C ) = 13° 58', 
j (c' + c) = 63° 58', 

£X d =67° 8', 



sin £ (c' + c) 

log cos =9.98697 
colog cos = 0.35764 
log cot = 9.62504 



cot i Xd. 



log sin =9.38266 
colog sin = 0.04646 
log cot = 9.62504 



log tan i(C" + C) = 9.96965 
£(C T/ + C) = 43° 0' 
±(C- C)= 6°28 / 



log tan i {€' 



C) = 9.05416 
■ C) - 6° 28'. 



C = S. 49° 28' E. = course from A. 
C = S. 36° 32' W. = course from B. 



To find the distance : 



cos j- (c 7 + c) . _ ^ 

cos J D = ~j- [ sin £ Xj. 

cos £ (c' — c) 



| (c' + c) = 63° 58', 
b(C'+ C) = 43° 0', 

i\ d =67° 8', 



£D = 56° 26'. 
D=112°52 / 



log cos 
colog COS 
log sin 



= 9.64236 
= 0.13587 
= 9.96445 



log cos i 2; = 9.74268 



: 6772 m. 



To find L of r. 
sin PV = sin & sin C. 
log sin e' =9.99021 
log sin C = 9.7747 3 
log sin PV= 9.76494 



To find X of V. 
cot CPY = cos c' tan C. 
log cos & =9.32025 
log tan C = 9.86974 
log cot CPr= 9.18999 



PF = 


35° 


36'. 


CPr= 81°12 / 


Xof F = 


90° 


/ -35°36 / 


XofC= 77° 14' 


= 


54° 


24 / S. 


Xof V= 158° 26' W. 


• 




By Rhumb Line : 




To find L d . 




To find X m . 


To find Xd. 


L' = 40° S. 




L' = 40° S. 


y = 148° 30' E. 


1" = 12° 4' S. 




L „ _ 12 o 4 / g 


X" = 77° 14' W. 


X d = 27° 56' 




2)52° 4' 


Xd = 225° 44' 


= 1676 m. 




X m = 26° 2 / 


= 360° - 225° 44 / 
= 134° 16' = 8056 m. 



360 



NAVIGATION. 



tanC 



To find the course. 

•* 

_ \d cos L m 
L d 

logX d = 3.90612 

logcos£ m = 9.95354 
colog L d = 6.77573 
log tan C = 10.63539 



To find the distance. 

.D = La sec C. 

log i d =3.22427 

log sec C = .646 82 

log 2) =3.87109 

1) - 7432 m. 



C = 76° 58'. That is, N. 76° 58' .E. from A, or S. 76° 58' W. from B. 



Exercise IX. Page 387. 



1. Observed altitude 
Correction . . 

True altitude 

2. Observed altitude 
Correction . . 

True altitude 

3. Observed altitude 
Correction . . 



True altitude 



25° 6' 10" f Index correction 

- 4' 50" J Dip 

I Refraction . . 

25° V 20" 

15° 20' 25" 

- 9' 44" . 

15° 10' 41" 

18° 17' 30" 
9' 41" 



Index correction 

Dip 

Refraction . . 



Index correction 

Dip 

Refraction . . 
Semi-diameter . 
Parallax . . . 



+ 1' 15" 

-4' 2" 
-2' 3.4' 



- 2' 20" 

- 3 / 55" 

- 3 / 29.4" 



+ o'i8" 

- 4' 9" 

- 2' 54" 
+ 16' 18" 

+ 8" 



18° 27' ll 7 



4. Observed altitude 
Correction . . 



True altitude 



30° 12' 40" r Semi-diameter . 
10' 24" I Parallax . . . 
i Index correction 
Dip . . . . . 
^ Refraction 
30° 23' 4" 



+ 16' 4" 
+ 8" 

- 0' 0" 

- 4' 16" 

- 1'39" 



5. Observed altitude 
Correction 



True altitude 



56° 25' 20" 
10' 18" 



56° 35' 28" 



" Semi-diameter . 
Parallax . . . 
Index correction 
Dip . . . . . 
Refraction 



4-16' 3" 
+ 5" 

- 1'20" 

- 4' 2" 

- 0' 38" 



TEACHERS EDITION. 



3G1 



6. Observed altitude . 


60° 10' 10" 


' Semi-diameter . 






+ 15' 48' 


Correction . . . 


9' 55" 


Parallax . . . 






+ 0' 4' 






Index correction 
Dip 






+ 2' 15' 

— 4' W 






-MSXfJ • • • . . 

Refraction . . 






- / 33' 


True altitude . . 


60° 19' 5" 










7. Observed altitude . 


31° 24 / 35" 


' Semi-diameter . 






+ 16' 14' 


Correction . . . 


10' 38" 


Parallax . . 






+ 8' 




< 


Index correction 

Dip 

Refraction . . 






- 0' 0' 

- 4' 9' 

- V 35' 


True altitude . . 


31° 35' 13" 










8. Observed altitude . 


26° 17' 20" 


' Semi-diameter . 






- 16' 10 


Correction . . . 


19' 53" 


Parallax . . 






+ 8' 




-< 


Index correction 
Dip . . . . 
Refraction . . 






+ 2' 15 

— 4' 9 

- P57 


True altitude . . 


25° 57' 27" 




9. Observed altitude . 


20° 35' 30" 


' Semi-diameter . 






- 15' 46 


Correction . . . 


21' 49" 


Parallax . . . 






+ 0' 8 






Index correction 
Dip 






+ 0' 18 
— 3' 55 






Refraction . . 






- 2' 34 


True altitude . . 


20° 13' 41" 










10. Observed altitude . 


36° 12' 10" 


' Semi-diameter . 






- 15' 47 


Correction . . . 


20' 57" 


Parallax . . 






+ 0' 8 






Index correction 

nip 

^ Refraction . . 






4- 0'25 

- 4' 23 

- P20 


True altitude . . 


35° 51' 13" 










Exercise X. 


Page 389. 








Astronomical 


Time. 


Civil Time 








d. h. 


m. s. 


d. h. 


m 


s. 




1. July 8 7 


10 = 


July 8 7 


6 


10 


P.M. 


2. Mar. 7 12 


25 30 = 


Mar. 8 


25 


30 


A.M. 


3. Jan. 1 18 


10 10 = 


Jan. 2 6 


10 


10 


A.M. 


4. Dec. 31 15 


= 


Jan. 1 3 








A.M. 


5. Feb. 2 8 


4 30 = 


Feb. 2 8 


4 


30 


P.M. 



362 



NAVIGATION. 



7. Mar. 

8. Aug. 

9. Sept. 
10. Jan. 



Civil Time. 

d. h. m. 



Astronomical Time. 



6. July 1 11 8 25 a.m. = June 30 23 8 25. 



11 

10 



10 



56 

8 

12 

41 



56 p.m. = Mar. 
20 p.m. = Aug. 
15 a.m. = Aug. 
56 a.m. == Dec. 



2 
31 
31 
31 



11 
10 
12 

22 



56 

8 

12 

41 



56. 
20. 
15. 
56. 



Exercise XI. Page 391. 







d. 


h. 


m. 


s. 




1. 


Ship date, 


May 4 


6 


12 


15 






Longitude in time, 




■f 11 


23 


20 


15)170°50 / 0" 




Greenwich date, 


May 4 


17 


35 


35 


11 h. 23 / 20 // 






d. 


h. 


m. 


s. 




2. 


Ship date, 


July 30 


23 


12 


30 






Longitude in time, 




+ 2 


41 


20 


15)40° 20' 0" 




Greenwich date, 


July 31 


1 


53 


50 


2h.41 / 20 // 






d. 


h. 


m. 


s. 




3. 


Ship date, 


July 31 


14 


10 


15 






Longitude in time, 




5 


22 


43 


15)80° 40' 45" 




Greenwich date, 


July 31 


19 


32 


58 


5 h. 22' 43" 






d. 


h. 


m. 


s 




4. 


Ship date, 


Mar. 2 


10 


20 









Longitude in time, 




3 


23 





15)50° 45' 




Greenwich date, 


Mar. 2 


6 


57 





3h.23 / 






d. 


h. 


m. 


s. 




5. 


Ship date, 


Mar. 25 


11 


8 


p, 


,M. 




Longitude in time, 




6 


41 


42 


15) 100° 25' 30" 




Greenwich date, 


Mar. 25 


17 


49 


42 


6 h. 41' 42" 






d. 


h. 


m. 


s. 




6. 


Greenwich date, 


Dec. 30 


19 


47 


28 






Longitude in time, 




1 


40 


28 


15)25° r 0" 






Dec. 30 


18 


7 





1 h. 40' 28" 




Local civil time, 


Dec. 31 


6 


7 


A 


,M. 






d. 


h. 


m. 


6. 




7. 


Greenwich date, 


July 4 


23 


51 









Longitude in time, 




11 


56 





15) 179° 0' 0" 




Local civil time, 


July 4 


11 


55 


p. 


m. 11 h. m' 



teachers' edition. 363 

d. h. DDL I. 

8. Greenwich date, July 3 23 59 
Longitude in time, 11 56 

July 3 35 55 
Local civil time, July 4 11 55 p.m. 

d. h. m. s. 

9. Greenwich date, May 19 19 40 20 
Longitude in time, 3 15 )45' 0" 

May 19 19 43 20 3' 

Local civil time, May 20 7 43 20 a.m. 

1880. d. h. m. s. 

10. Greenwich date, Dec. 31 15 8 

Longitude in time, 8 40 15 )2° 10' 0" 

Dec. 31 15 16 40 8' 40" 

Local civil time, Jan. 1 3 16 40 a.m. 



Exercise XII. Page 396. 

Find the sun's declination and the equation of time corresponding to 
the following Greenwich dates : 



1. 1895 Jan. 7 d. 3 h. apparent time. 

m. 8. 

Jan. 7 d. h. 0's dec. 22° 22' 25.8" S. Eq. of time + 6 29.57 

Diff. for3h. -57.7" + 3.22 



Jan. 7d. 3h. 0's dec. 22°21 / 28.1 // S. Eq. of time + 6 32.79 

2. 1895 Aug. Id. 6h. 12 m. 20s. apparent time. 

m. a. 

Aug. 1 d. h. 0's dec. 18° V 59.0" N. Eq. of time + 6 7.87 

Diff. for 6 h. 12 m. 20 s. - 3' 54.3" - 0.92 



Aug. 1 d. 6 h. 12 m. 20 s. O's dec. 17° 58' 4.7" N. Eq. of time + 6 6.95 



3. 1895 May 5 d. 10 h. 25 m. apparent time. 



m. 6. 



May 5 d. Oh. 0's dec. 16° 15' 42.5" N. Eq. of time - 3 26.15 

Diff. for 10 h. 25 m. + 7 / 25.4" - 2.33 



May 5 d. 10 h. 25 m. 0's dec. 16° 23' 7.9" N. Eq. of time - 3 28.48 



364 NAVIGATION. 

4. 1895 Aug. 7d. 15 h. 12 m. apparent time. 

m. g. 
Aug. 8 d. h. O's dec. 16° 9' 20.0" N. Eq. of time + 5 28.13 

Diff. for 8 h. 48 m. + 6 / 15.0" + 2.84 



Aug. 8 d. 15 h. 12 m. O's dec. 16° 15' 35.0" N. Eq. of time + 5 30.97 
5. 1895 Dec. 4d. 6h. 18 m. apparent time. 

m. s. 

Dec. 4d. Oh. O's dec. 22° 15' 34.0" S. Eq. of time - 9 40.21 

Diff. for 6 h. 18 m. -f 2' 6.7" + 6.38 



Dec. 4d. 6h. 18 m. O's dec. 22° 17' 40.7" S. Eq. of time - 9 33.83 

6. 1895 July 23 d. 20 h. 16 m. 40 s. apparent time. 



m. s. 



July 24 d. Oh. O's dec. 19° 53' 2.9" N. Eq. of time + 6 16.42 

Diff. for 3 h. 43 m. 20 s. + i'57.3" - 2.12 



July 23d. 20 h. 16m. 40s. O's dec. 19° 55' 0.2" N. Eq. of time + 6 14.30 



7. 1895 Nov. Id. 3h. 6m. apparent time. 



m. 8. 



Nov. 1 d. h. O's dec. 14° 26' 57.9" S. Eq. of time - 16 18.64 

Diff. for 3 h. 6 m. + 2' 29.0" - 0.20 



Nov. 1 d. 3 h. 6 m. O's dec. 14° 29' 26.9" S. Eq. of time - 16 18.84 

8. 1895 Oct. 12 d. 5h. 12 m. apparent time. 

m. s. 

Oct. 12 d. Oh. O's dec. 7° 24' 29.3" S. Eq. of time - 13 26.96 

Diff. for 5 h. 12 m. + 4' 53.4" - 3.20 



* Oct. 12 d. 5 h. 12 m. O's dec. 7°30 / 22.7"S. Eq. of time - 13 30.16 

9. 1895 June 7d. 3h. 18m. meantime. 

m. b. 

June 7 d. h. O's dec. 22° 45' 50.5" N. Eq. of time -f 1 26.59 

Diff. for 3 h. 18 m. + 47.9" - 1.51 



June 7 d. 3 h. 18 m. O's dec. 22° 46' 38.4" N. Eq. of time + 1 25.08 

10. 1895 Feb. 3d. 9h. 15m. meantime. 

m. s. 

Feb. 3 d. h. O's dec. 16° 30' 23.8" S. Eq. of time - 14 2.48 

Diff. for 9 h. 15 m. - 6' 48.9" -2.40 



Feb. 3d. 9h. 15m. O's dec. 16°23 / 34.9" S. Eq. of time - 14 4.88 



TEACHERS 7 EDITION. 



365 



Exercise XIII. Page 404. 

1. Given civil date 1895 Jan. 1, longitude 102° 41' W., observed 
meridian altitude of 59° 59' 50" S. , index correction -f 0' 50", eye 
15 ft. ; find the latitude. 



Long. 102° 4r W .= 6h. 50 m. 44 s. 

© 59° 59' 50" f Index cor., + 0' 50" 
Semi-diam., + 16' 18" 



+ 12' 50" 



Dip, 



3' 48" 



Refraction, — 0'34" 
Parallax, + 0' 4" 



O'sdec. 23° 0' 34" S. ! 12.49 



1' 25" 



d = 22° 59' 9" S. 
z = 29° 47' 20" X. 
L= 6°48 / ll // N. 



6.84 



85.43 



60° 12' 40" 

90°^ 

z = 29° 47' 20" N. 



2. Given civil date 1895 Feb. 1, longitude 78° 14' E., observed meridian 
altitude of 78° 4' 10" S., index correction + 0' 55", eye 12 ft.; find the 
latitude. 

Long. 78° 14' = 5 h. 12 m. 56 s. 



78° 4' 10" 


'Index cor., + 0' 55" 
Semi-diam., + 16' 16" 
Dip, - 3' 24" 
Refraction, — 0' 12" 

^ Parallax, + 0' 2" 


O's dec. 17° 5' 1" S. 
3' 43" 


42.76 
5.22 


+ 13' 37" - 


d = 17° 8' 44" S. 
z = 11° 42' 13" N. 

L= 5° 26' 31" S. 


223.21 


78° 17' 47" 
90° 







z = 11° 42' 13" N. 



3. Given civil date 1895 Mar. 20, longitude 173° 18' W., observed 
meridian altitude of 89° 37' 0" N., index correction + 4' 32", eye 18 ft. ; 
find the latitude. 



Long. 173° 18' = ] 


Llh. 33 m. 12 s. 






O 89° 37' 0"N. 


r Index cor., + 4' 32" 


O'sdec. 0° 8'36"S. 


59.26 




Semi-diam., + 16' 5" 
Dip, - 4' 9" 


- 11' 24" 


11.55 


4- 16' 28" «| 


d = 0° 2'48"N. 


684.45 




Refraction, — 0' 0" 
^ Parallax, + 0' 0" 


z = 0° 6'32"S. 




i = 0° 3'44"S. 


89° 53' 28" 






90° 









z= 0° 6'32"S. 



366 



NAVIGATION. 



0'sdec. 4° 33' 23" N. 


57.85 


+ 5' 38" 


5.85 


d= 4° 39' 1"N. 


338.42 


z = 41° 4-4o^n. 




L = 45° 43' 41" N. 





4. Given civil date 1895 April 1, longitude 87° 42' W., observed merid- 
ian altitude of 48° 42' 30" S., index correction + T 42", eye 18 ft.; 
find the latitude. 

Lo ng. 87° 42 7 - 5h. 50 m. 48 s. 

48° 42' 30" S. f Index cor. , + 1' 42" 

| Semi-diam., + 16 7 2" 

+ 12' 50" -j Dip, - 4 / 9" 

Refraction, — 0' 6V' 

Parallax, + 0' 6" 

48° 55 / 20" S. 
90° 
2=41° 4'40"N. 

5. Given civil date 1895 Sept. 1, longitude 97° 42' E., observed merid- 
ian altitude of © 51° 4' 50" S. , index correction — 6' 0", eye 23 ft. ; find 
the latitude. 

Lo ng. 97°42 / = 6h. 30 m. 48 s. 

Index cor., — 6' 0" 
Semi-diam., + 15' 54" 
Dip, - 4' 42" 

Refraction, — 0'47" 
^Parallax, + 0' 



51° 4' 50" S, 
+ 4' 31" 



0'sdec. 8°17 / 14"N. 


54.43 


5' 54" 


6.51 


d = 8° 23' 8"N. 


354.34 


z = 38° 50' 39" N. 




Z, = 47° 13' 47" N. 





51° 9 / 21"S. 
90^ 

2 = 38° 50' 39" N. 



6. Given civil date 1895 Aug. 26, longitude 92° 3' E., observed meridian 
altitude of © 35° 35' 20" N., index correction + 2' 17", eye 12 ft.; find 
the latitude. 

Lo ng. 92° 3' = 6 h. 8 in. 12 s. 

35°35 / 20"N. f Index cor. , + 2' 17" 

Semi-diam., -f 15' 52" 

+ 13' 31" < Dip, - 3' 24" 

Refraction, — l'2'l" 

Parallax, + 0' 7" 

35° 48' 51" N. 

9CP 

z- 54° ll 7 9"S. 



0'sdec. 10°25 / 18"N. 


52.22 


5' 21" 


6.14 


d =10° 30' 39" N. 


320.66 


z = 54°ll' 9"S. 




L = 43° 40' 30" S. 





7. Given civil date 1895 May 16, longitude 45° 26" W., observed merid- 
ian altitude of 86° 34' 20" N., index correction + 4' 16", eye 15 ft.; 
find the latitude. 






TEACHERS EDITION. 



367 



Long. 45° 26' = 3 h. 1 m. 44 s. 








86° 34' 20" N. 


' Index cor. , 


+ 4' 16" 


0'sdec. 19° 6'29"N. 


34.63 




Semi-diam. 
Dip, 


+ 15' 51" 

- 3' 48" 


V 45" 


3.03 


+ 16' 16" \ 


d = 19° 8' 74" N. 


104.93 




Refraction, 


- 0' 4" 


z= 3° 9'24"S. 




^ Parallax, 


+ 0' 1" 


L =15° 58' 50" N. 


86° 50' 36" N. 






90° 











z= 3° 9' 24" S. 



8. Given civil date 1895 March 20, longitude 174° 0' W., observed 
meridian altitude of 89° 56' 10" N., index correction — 1' 15", eye 
15 ft. ; find the latitude. 

Lo ng. 174° 0'= 11 h. 36 m. 

89° 56' 10" N. f Index cor., - 1' 15" O's dec. 0° 8' 36" S. 59.26 



I Semi-diam., + 16' 5" 

+ 10' 52" \ Dip, - 3' 48" 

Refraction, — 0' 0" 

{ Parallax, + 0' 0" 

90° 7' 2"N. 
9CP 

z = o° 7' 2"N. 



11' 27 



d = 0° 2'51"N. 

z = qq y 2"N. 
L = 0° 9'53"N. 



11.60 



687.42 



9. Given civil date 1895 June 1, longitude 44° 40' E., observed meridian 
altitude of 72° 14' 10" N., index correction + 3' 45", eye 22 ft.; find 
the latitude. 



Long. 44° 40' = 2 h. 58 in. 40 s. 






72° 14' 10" N. 


r Index cor., + 3' 45" 


0'sdec. 22° 3'54"N. 


20.39 




Semi-diam., -f 15' 48" 
Dip, - 4' 36" 


1' 1" 


2.98 


+ 14' 41" -! 


d = 22° 2'53"N. 


60.76 




Refraction, — 0' 19" 


z = 17° 31' 9" S. 






Parallax, + 0' 3" 


i= 4° 31' 44" N. 




72° 28' 51" N. 






90° 








2=17° 31' 9"S. 





10. Given civil date 1895 Dec. 1, longitude 67° 56' E. , observed meridian 
altitude of 18° 48' 10" S., index correction — 3' 6", eye 18 ft.; find the 
latitude. 



368 



NAVIGATION. 



Lo ng. 67° 56' = 4 h. 31 m. 44 s. 



18° 48' 10" S. 


' Index cor. , 


- 3' 6" 


0'sdec. 21° 49' 31" S. 




Semi-diam. 
Dip, 


+ 16' 16" 
- 4' 9" 


V 45" 


4- 6' 20" J 


d = 21° 51' 16" S. 




Refraction, 


- 2' 49" 


2 = 71° 5'30"N. 




Parallax, 


+ 0' 8" 


£ = 49° 14' 14" N. 


18° 54' 30" S. 






90° 






z =7i° 6'30"N. 









23.29 

4.53 

105.04 



11. Given civil date 1895 Sept. 23, longitude 57° 45' E., observed 
meridian altitude of 84° 10' 50" N. , index correction — V 36", eye 
16 ft. ; find the latitude. 

Lo ng. 57° 45' .= 3 h. 51 m. 



84° 10' 50" N. 


'Index cor., - 1'36" 
Semi-diam., +15' 59" 
Dip, - 3' 55" 
Refraction, — 0' 6" 
Parallax, + 0' 1" 


0'sdec. 0° 4 / 35"S. 
3' 45" 


58.49 
3.85 


+ 10' 23" - 


d = 0° 0'50"S. 

z = 5° 38' 47" S. 
L = 5° 39' 37" S. 


225.19 


84° 21' 13" N. 
90° 






z= 5° 38' 47" S. 





12. Given civil date 1895 Sept. 23, longitude 119° 54' E., observed 
meridian altitude of 83° 46' 0" S. , index correction — 5' 30", eye 
18 ft. ; find the latitude. 



Long. 119° 54' = 


7h. 59 m. 36 s. 




83° 46' 0"S. 


r Index cor., - 5' 30" 


0'sdec. 0° 4 / 35"S. 


58.49 




Semi-diam., + 15' 59" 
Dip, - 4' 9" 


r 48" 


7.99 


+ 6' 15" J 


d = 0° 3'13"K 


467.60 




Refraction, — 0' 6" 


z = 6° 7'45"N. 




Parallax, + 0' 1" 


£ = 6°10'58"N. 


83° 52' 15" S. 






90° 






z- 6° 7'45"N. 





13. Given civil date 1895 Nov. 21, longitude 70° 20' E., observed 
meridian altitude of 80° 20' 0" N., index correction — 2' 50", eye 20 ft. ; 
find the latitude. 



TEACHERS' EDITION. 



369 



Lo ng. 70° 20'- 4 h. 41m. 20 s. 



80° 20' 0"N. 


' Index cor. , — 2' 50" 


O's dec. 19° 56' 16" S. 


33.11 




Semi-diam., + 16' 14" 
Dip, - 4' 23" 


2' 35" 


4.69 


+ 8' 53" «) 


d - 19° 53' 31" S. 


155.29 




Refraction, — 0' 10" 


2=9° 31' 7" S. 






Parallax, + 0' 2" 


L = 29° 24' 38" S. 




80° 28' 53" N. 




90° 









9° 31' 7"S. 



14. Given civil date 1895 Dec. 31, longitude 123° 45' W., observed 
meridian altitude of 67° 8' 10" 8., index correction -f 0' 9", eye 13 ft.; 
find the latitude. 



Lo ng. 123° 45' = 8 h. 15 m. 



67° 8' 10" S. 


' Index cor. , 


+ 0' 9" 


O's dec. 23° 6' 22" 


S. 


11.03 




Semi-diam. 
Dip, 


+ 16' 18" 
- 3' 32" 


1'31' 




8.25 


+ 12' 34" «| 


d = 23° 4' 51' 


s. 


91.00 




Refraction, 


- 0'25" 


z = 22° 39' 16" 


N. 






Parallax, 


+ 0' 4" 


X= 0°25'35" 


S. 




67° 20' 44" S. 








90° 













z=22°39'16"N. 



15. Given civil date 1895 Oct. 20, longitude 150° 25' W., observed 
meridian altitude of 49° 58' 50" N., index correction + 1' 10" eye 
19 ft. ; find the latitude. 

Lo ng. 150° 25' = 10 h. 1 m. 40 s. 



49° 58' 50" 


N. 


r Index cor. , 


+ l'lO" 


O 


s dec. 10° 21' 17" 


S. 


53.89 




* 


Semi-diam. 
Dip, 


-f 16' 7" 
- 4' 16" 




9' 11' 




10.03 


+ 12' 18' 


d = 10° 30' 18" 


s. 


540.52 






Refraction, 


- 0'49" 




z = 39° 48' 52" 


s. 








Parallax, 


+ 0' 6" 




L = 50° 19' 10" 


s. 




50° 11° 8' 


'N. 




90° 
















z = 39° 48' 52' 


'S. 





16. Given civil date 1895 June 1, longitude 96° 17' E., observed merid- 
ian altitude of 75° 38' 15" N., index correction + 0' 27", eye 26 ft.; 
find the latitude. 



370 



NAVIGATION. 



Lo ng. 96° 17' = 6 h. 25 m: 8 s. 



75° 38' 15" N. 


' Index cor. , 
Semi-diam. 
Dip, 

Refraction, 
Parallax, 


+ 0'27" 
+ 15' 48" 

- 5' 0" 

- 0'15" 
+ 0' 2" 


O'sdec. 22° 3'54"N. 

2' 11" 


20.39 
6.42 


+ 11' 2" - 


d = 22° 1'43"N. 
z = 140 w 43- s 

L= 7° 51' 0"N. 


130.90 


75°49'17"N. 
90° 







z= 14° 10' 43" S. 

17. Given civil date 1895 June 25, longitude 59° 15' E., observed 
meridian altitude of 60° 23' 15" N., index correction + 2' 21", eye 
30 ft. ; find the latitude. 
Long L 59°15' = 31i. 57 m. 



O 60° 23' 15" N. 


r Index cor., + 2' 21" 


O'sdec. 23° 24' 19" N. 


3.93 




Semi-diam., — 15' 46" 
Dip, - 5' 22" 


15" 


3.95 


- 19' 16" J 


d = 23° 24' 34" N. 


15.52 




Refraction, — 0'33" 


z = 29° 56' 1"S. 






L Parallax, + 0' 4" 


L- 6° 31' 27" S. 




60° 3'59"N. 




90° 








2=29° 56' 1"S. 





Exercise XIV. Page 405. 

1. Given civil date 1895 Jan. 29, observed meridian altitude of Aldeb- 
aran 52° 36' 0" S., index correction — 0' 23", eye 20 ft.; find the latitude. 
Obs. alt. = 52° 36' 0" S. 

- 5' 31' 
True alt. 



52° 30' 29" S. 

9(P K 

Zenith dis. = 37° 29' 31" N. 
Dec. = 16° 18' 2"N. 

Lat. = 53° 37' 33" N. 



Index correction, — 23" 
Dip, - 4' 23" 

Refraction, — 45'' 



5' 31' 



2. Given civil date 1895 Feb. 18, observed meridian altitude of Procyon 



77° 18' 10" S., index correction + 0' 19" 
Obs. alt. = 77° 18' 10" S. 
- 3' 50" 



True alt. 


= 77° 
90° 


14' 


20' 


'S. 

N. 


Zenith dis. 


= 12° 


45' 


40' 


'N. 


Dec. 


= 5° 


29' 


39' 


'N. 


Lat. 


= 18° 


15' 


19' 


'N. 



eye 16 ft. ; find the latitude. 

Index correction, + 19" 
Dip, - 3' 55" 

Refraction, — 1' 13.5" 

- 3' 49.5" 



TEACHERS' EDITION. 



371 



3. Given civil date 1895 March 20, observed meridian altitude of 
Arcturus 36° 10' 20" N., index correction + 2' 42", eye 20 ft.; find the 
latitude. 



Obs. alt. = 36° 10' 20" N. 
-3' 1" 



True alt. 


= 36° 
90° 


7' 


19' 


'N. 
S. 


Zenith dis. 


= 53° 


52' 


41" 


'S. 


Dec. 


= 19° 


43' 


23' 


'N. 


Lat. 


= 34° 


9' 


18" 


'S. 



Index correction, -f 2' 42" 
Dip, - 4' 23" 

Refraction, — 1' 20" 



-3' 1' 



4. Given civil date 1895 Aug. 17, observed meridian altitude of Altair 
66° 51' 10" N., index correction + 0' 58", eye 13 ft.; find the latitude. 



Obs. alt. = 66° 51' 10" N. 

-3' 0" 
True alt. = 66° 48' 10" N. 

90°^ & 

Zenith dis. = 23° 11' 50" S. 
Dec. = 8° 35' 34" N. 

Lat. = 14° 36' 16" S. 



Index correction, -+- 58" 
Dip, - 3' 32" 

Refraction, — 25.5" 



-3' 0" 



5. Given civil date 1895 Nov. 4, observed meridian altitude of Fomal- 
haut 59° 40' 0" N., index correction + V 12", eye 23 ft. ; find the latitude. 

Obs. alt. =59° 40' 0"N. 
-4' 4" 



True alt. 



: 59° 35' 56" N. 
90° S. 



Zenith dis. = 30° 24' 4" 
Dec. = 30° 10' 32" 

Lat. 



: 60° 34' 36" S. 



Index correction, 1' 12" 
Dip, — 4' 42" 

Refraction, — 34" 



-4' 4" 



6. Given civil date 1895 Sept. 6, observed meridian altitude of Arcturus 
86° 35' 50" N., index correction — V 10", eye 12 ft.; find the latitude. 



Index correction, — V 10" 
Dip, - 3' 24" 

Refraction, — ; 4^ 

- 4' 38" 



Obs. alt. 


= 86° 


35' 


50" 








-4' 


38" 




True alt. 


= 86° 
= 90° 


31' 


12" 


S. 


Zenith dis, 


= 3° 


28' 


48" 


s. 


Dec. 


= 19° 


43' 


37" 


'N. 


Lat. 


= 16° 


14' 


49" 


'N. 



372 NAVIGATION. 

7. Given civil date 1895 Oct. 6, observed meridian altitude of Markab 
54° 10' 15" S. , index correction 0, eye 13 ft. ; find the latitude. 

•Obs. alt. = 54° 10' 15" S. 

— 4' 14" 

Index correction, -f 0' 0" 

Dip, — 3" 32" 

Eefraction, — 42" 

- 4' 14" 



8. Given civil date 1895 Aug. 17, observed meridian altitude of /3 Cen- 
tauri 59° 47 / 13" S. , index correction 0, eye 25 ft. ; find the latitude. 

Obs. alt. = 59° 47' 13" S. 



True alt. 


= 54° 
= 90° 


6' 


V 


'S. 

N. 


Zenith dis. 


= 35° 


53' 


59'' 


'N. 


Dec. 


= 14° 


38' 


49" N. 


Lat. 


= 50° 


32' 


48"N. 



5' 28" 



True alt. 


= 59° 
= 90° 


41' 


45" 


'S. 

N. 


Zenith dis. 


= 30° 


18' 


15" 


'N. 


Dec. 


= 59° 


52' 


29" 


'S. 


Lat. 


= 29° 


34 / 


14" 


'S. 



Index correction, -f 0' 0" 
Dip, - 4' 54" 

Refraction, — 34" 



• 5' 28" 



9. Given civil date 1895 Dec. 4, observed meridian altitude of a Arietis 
60° 29' 50" S., index correction - 2' 10", eye 18 ft.; find the latitude. 

Obs. alt. =60°29 / 50"S. 
- 6' 52" 



True alt. 


= 60° 


22' 


58" S. 


Index correction, 


-2' 


10" 




90° 






N. 


Dip, 

Refraction, 


— 4' 


9" 
33" 


Zenith dis. 


= 29° 


:rr 


2"N 




















— 6' 


M" 


Dec. 


= 22° 


58' 


26" 


'N. 








Lat. 


= 52° 


35' 


28" 


'N. 









10. Given civil date 1895 Feb. 8, observed meridian altitude of Sirius 
37° 50' 20" S., index correction -f V 4", eye 19 ft.; find the latitude. 

Obs. alt. = 37° 50' 20" S. 



- 4 / 27" 
True alt. = 37° 45' 53" S. 

90° N. 



Zenith dis. = 52° 14' 7"N. 
Dec. = 16° 34 7 20" S. 

Lat. = 35° 41' 47" N. 



Index correction, -J- V 4" 
Dip, - 4' 16" 

Refraction, — V 15" 

- 4' 27" 



TEACHERS EDITION. 



373 



11. Given civil date 1895 April 9. observed meridian altitude of Sirius 

61° 3' 50" X., index correction 0, eye 16 ft. ; find the latitude. 

Obs. alt. =61° 3'50"X. 

4' 27" 

Index correction, + 0' 0" 

Dip, * -3' 55" 

Refraction, — 32" 



True alt. 


= 60° 
90° 


59 / 


23" 


X. 

s. 


Zenith dis. 


= 29° 


0' 


37' 


'S. 


Dec. 


16° 


34' 


24* 


s. 


Lat. 


= 45° 


35' 


r 


s. 



4' 27 



12. Given civil date 1895 March 30, observed meridian altitude- of Spica 
52° 14' 0" N., index correction 0, eye 19 ft.; find the latitude. 
Obs. alt. = 52° 14 / 0" N. 
5' 1" 



True alt. =52° 8'59"X. 

90°^ S. 

Zenith dis. = 37° 51' 1" S. 
Dec. = 10° 37 / 4" S. 

Lat. = 48° 28' 5" S. 



Index correction. + 0' 0" 

Dip, - 4' 16" 

Refraction, — 45" 

— 5' 1" 



13. Given civil date 1895 July 8, observed meridian altitude of Antares 
70° 10' 30" X.. index correction 0, eye 21 ft.; find the latitude. 
Obs. alt. = 70° 10' 30" N. 

Index correction, 4- 0' 0" 

Dip, - 4' 29" 

Refraction, — 21" 

- 4 / 50" 



True alt. 


= 70° 
90° 





40" 


N. 

S. 


Zenith dis 


= 19° 


or 


20" 


S. 


Dec. 


= 26° 


12' 


12" 


s. 


Lat. 


= 46° 


6' 


32" 


s. 



Exercise XY. Page 413. 

1. 1895. Oct. 19. a.m.. at sea. in latitude 33° 27' S. ; the observed alti- 
tude 28° 22' 30" ; index correction + 30" ; height of eye 18 ft. : Green- 
wich mean time by chronometer Oct. 18 d. 18 h. 28 m. 38 s. Required 
the longitude. 



28°22 / 30" 


' Index cor. , 


+ O^O" 


I 


sdec. 9° 59' 52.4" S. 


54.27 




Semi-diam. 


+ 16' 6" 




4 / 59.6" 


5.52 


+ 10' 47", 


Dip, 


- 4' 9" 




9- 54' 52.8" S. 


299.57 




Refraction, 


— 1 , 48 ,/ 




90° S. 






Parallax, 


+ 0' 8" 




p = 80° 5' 7.2" S. 





h= 28°33 / 17" 



374 



NAVIGATION. 



h- 28° 33' 17" 
L= 33° 27' 0" 
p= 80° 5' 7" 



2S= 142° 5' 24" 



22 = 



71° 2' 42" 
42° 29' 25" 



log sec = 
log esc = 

log cos = 
log sin 



0.07864 
0.00654 

9.51165 

9.82960 



Oct. 



2 )19.42643 
logsin££ = 9.71321 
i£ = 31°6'31 

d. h. m. s. 

18 19 51 8 
14 54 



Equation of time. 

m. s. 

14 56.76 

2.47 

14 54.29 



0.447 
5.52 



2.47, 



Oct. 
Oct. 



t = 62° 13' 2" 



4h. 8m. 52s. 



19 36 

18 28 



14 

38 



1 7 36 
16° 54' 0" 



Local apparent astronomical time. 

Equation of time. 

Local mean astronomical time. 

Greenwich mean time. 

Difference of time. 

E. Longitude. 



2. 1895, Oct. 20 a.m., at sea, in latitude 31° 40' S. ; the observed alti- 
tude 35° 16' 10" ; index correction + 30" ; height of eye 18 ft. ; Green- 
wich mean time by chronometer, Oct. 19 d. 19 h. 11m. 24 s. Kequired 
the longitude. 



35° 16' 10" 



+ 11' 14" 



h = 


35° 27' 24" 


L = 


31° 40' 0" 


P = 


79° 42' 49" 



2 S = 146° 50' 13" 
8= 73° 25' 6" 
R= 37° 57' 42" 



Index cor. , 
Semi-diam. , 
Dip, 

Eefraction, 
Parallax, 

log sec = 
log esc = 

log cos = 
log sin = 



+ 


0' 


30" 


+ 16' 


7" 


— 


4' 


9" 


— 


V 


22" 


+ 


0' 


8" 



0'sdec. 10° 21' 30.4" 
4' 19.2" 



S. 



10° 17' 11' 

90° 



53.89 
4.81 



259.21 



0.07001 
0.00704 

9.45543 

9.78897 



p =z 790 42' 49" S. 
Equation of time. 



15 7.17 

2.03 

15 5.14 



0.421 
4.81 



2.03 



Oct. 



2 )19.32145 
logsini*= 9.66072 
i£ = 27°14'53 

d. h. m. e. 

19 20 22 1 
15 5 



t = 54° 29' 46" = 3 h. 37 m. 59 s. 



Oct. 19 20 6 56 
Oct. 19 19 11 24 



55 32 
13° 53' 0" 



Local apparent astronomical time. 

Equation of time. 

Local mean astronomical time. 

Greenwich mean time. 

Difference in time. 

E. Longitude. 



TEACHERS' EDITION. 



375 



3. 1895, Oct. 20, p.m., at sea, in latitude 30° 55' S.; the observed alti- 
tude 21° 42" 30" ; index correction + 29" ; height of eye 18 ft. ; Green- 
wich mean time by chronometer Oct. 20 d. 3h. 35 m. 40 s. Required the 
longitude. 



21° 42' 30" 



+ w ir 



21° 52' 41" 
30° 55' 0" 

79° 35' 16" 



h = 
L = 

P = 

2 5=132° 22' 57 
8= 66° 11' 29 

R= 44° 18' 48 



f Index cor., + 0' 29" 
Semi-diam., -+- 16' 7" 
Dip, - 4' 9" 

Refraction, — 2' 24" 

L Parallax, + 0' 8" 



dec. 



10° 21' 30.4" 
3' 13.5" 



S. 



10° 24' 44" 

90° 



53.89 
3.59 



193.47 



log sec = 
log esc = 

log cos = 
log sin = 



0.06656 
0.00721 

9.60604 
9.84421 



p = 79° 35' 16" S. 
Equation of time. 

m. s. 

15 7.17 
1.51 

15 8.68 



0.421 
3.59 



1.51 



2 )19.52402 
logsin££ = 9.76201 
££=35° 19' 3" 



t = 70° 38' 6" = 4 h. 42 m. 32 s. 



d. 

Oct. 20 



Oct. 20 
Oct. 20 



4 42 32 

15 9 

4 27 23 

3 35 40 



51 43 
12° 55' 45" 



Local apparent astronomical time. 

Equation of time. 

Local mean astronomical time. 

Greenwich mean time. 

Difference in time. 

E. Longitude. 

4. 1895, Oct. 21, a.m., at sea, in latitude 29° 35' 8.; the observed alti- 
tude 24° 26' 42" ; index correction + 29" ; height of eye 18 ft. ; Green- 
wich mean time by chronometer Oct. 20 d. 18 h. 30 m. 39 s. Required 
the longitude. 

-f 0'29 

+ 16' 7 

- 4' 

- 2' 
+ 0' 



24° 26' 42" 



+ 10'27 



h = 


24° 37' 9" 


L = 


29° 35' 0" 


P = 


79° 21' 54" 



f Index cor. , 
Semi-diam. , 
Dip, 

Refraction, 
Parallax, 



9 



8' 



0'sdec. 10° 42' 59.3" 
4' 53.7" 



S. 



10° 38' 
90° 



6" 



53.50 
5.49 



293.71 



log sec = 
log esc = 



0.06066 
0.00752 



2 8=133° 34' 3' 
S = 66° 47' T 
R= 42° 9 / 62' 



' log cos = 9.59573 

' log sin = 9.82691 

2 )19.49082 

logsin££= 9.74541 



p = 79° 21' 54" S. 
Equation of time. 

m. s. 

15 16.94 

2.16 

15 14.78 



0.394 
5.49 



2.16 



i t = 33° 48' 33". t = 67° 37' 6" = 4 h. 80 m. 28 s. 



376 





d. h. m. s. 


HiJX\ lUAllUi?l . 


Oct. 


20 19 29 32 


Local apparent astronomical time. 




15 15 


Equation of time. 


Oct. 


20 19 14 17 


Local mean astronomical time. 


Oct. 


20 18 30 39 


Greenwich mean time. 




43 38 


Difference in time. 




10° 54' 30" 


E. Longitude. 



5. 1895 Jan. 29, p.m., at ship, latitude 42° 26" N.; observed altitude 
13° 40' ; index error — V 8" ; height of eye 16 ft. ; time by chronometer 
29 d. 6 h. 48 m. 40 s., which was slow 11 m. 22.3 s. for mean noon at 
Greenwich, Dec. 1, 1894, and on Jan. 1, 1895, was 8 m. 7 s. slow for 
Greenwich mean noon. Required the longitude. 

Chronometer. 

m. s. d. h. m. «. 



1894 Dec. 

1895 Jan. 



1 slow 11 22.3 

1 slow 8 7.0 

31 ) 3 15.3 

6.3 

28.28 



Jan. 29 6 48 40 
+ 8 7 
-2 58 



2 58.2 



13° 40' 0" 

+ 7' 27" 



h= 13° 47' 27' 
L= 42° 26' 
p - 1Q7° 50' 28 
2 S = 164° 3' 55 
<S= 82° 1'57 
B= 68° 14' 30 



' Index cor., — V 8' 
Semi-dlam., + 16' 16' 
Dip, - 3 / 55 / 

Refraction, — 3' 55' 
Parallax, + 0' 9' 

log sec = 0.13191 
log esc = 0.02141 



Jan. 29 6 53 49 




O'sdec. 17°55 / 7.1" S. 


40.43 


4' 39.0" 


6.90 


17° 50' 28" S. 


278.97 


90° N. 





log cos = 9.14180 

log sin = 9.96790 

2)19.26302 

log sin J* = 9.63151 

J* = 25° 20' 42" 

d. h. m. s. 



p= 107° 50' 28" N. 
Equation of time. 

m. s. 

13 20.81 

3.10 

13 23.91 



0.435 
6.90 



3.10 



t = 50° 4r 24" = 3 h. 22 m. 46 s. 



Jan. 29 3 22 46 

13 24 

Jan. 29 3 36 10 

Jan. 29 6 53 49 

3 17 39 

49° 24' 45" 



Local apparent astronomical time. 

Equation of time. 

Local mean astronomical time. 

Greenwich mean time. 

Difference in time. 

W. Longitude. 



teachers' edition. 



377 



6. 1895, March 31, a.m., at ship, latitude 26° 9' N. ; observed altitude 
29° 10' 20"; height of eye 26 ft.; time by chronometer 31 d. Oh. 4m. 
50 s., which was 58 m. 58 s. fast for mean noon at Greenwich, Nov. 20, 
1894, and on December 31, 1894, was lh. 2 m. 55.8 s. fast for mean time 
at Greenwich. Required the longitude. 



Chronometer. 

h. m. s. 

Nov. 20 fast 58 58 
Dec. 31 fast 1 2 55.8 
41 ) 3 57.8 
5.8 
90. 



d. h. m. s. 

Mar. 31 4 50 
-1 2 56 
- 8 42 

Mar. 30 22 53 12 



8 42.0 



© 29° 10' 20" r Index cor., + 0' 0" O's dec. 4° 10' 8.3" N. 58.06 



+ 9' 26" 



h = 


29° 19' 46" 


L = 


26° 9' 0" 


P = 


85° 50' 57" 



Semi-diam., + 16' 2' 
Dip, - 5' 0' 

Refraction, — 1'44 A 
Parallax, + 0' 8' 

log sec = 0.04690 
log esc = 0.00114 



V 5.6" 



4° 
90° 



9' 3" 



N. 
N. 



2S= 141° 19' 43' 
S = 70° 39' 51' 
R= 41° 20' 5' 



4 15.13 

0.86 
4 15.99 



' log cos = 9.51996 

' log sin = 9.81984 

2 )19.38784 

logsini£= 9.69392 

££=29° 37' 5". 

t = 59° 14' 10" = 3 h. 56 m. 57 s. 



Local apparent astronomical time. 

Equation of time. 

Local mean astronomical time. 

Greenwich mean time. 

Difference in time. 

W. Longitude. 



Mar. 


d. h. m. e. 

30 20 3 3 
4 16 


Mar. 
Mar. 


30 20 7 19 
30 22 53 12 




2 45 53 
41° 28 / 15" 



1.13 



65.61 



p = 85° 50' 57" N. 
Equation of time. 



0.758 
1.13 



0.86 



7. 1895, May 22, a.m., at ship, latitude 43°25 / N.; observed altitude 
32° 8' ; index correction -f V 28" ; height of eye 15 f t. ; time by chro- 
nometer 21 d. 21 h. 6 m. 10 s., which was slow 12.6 s. for mean noon at 
Greenwich, Feb. 24, and on April 1 was 2 m. 45 s. fast for mean noon at 
Greenwich. Required the longitude. 



378 



NAVIGATION. 



Chronometer. 

m. b. 

Feb. 24 slow 12.6 
Apr. 1 fast 2 45.0 



32° 8' 0" 



+ 12' h"< 



36 )2 57.6 

4.93 

51 

4 11.4 

Index cor., + 1'28' 

Semi-diam., + 15' 50' 

Dip, - 3' 48' 

Refraction, — Y 33' 

Parallax, + 0" 8 / 



0's dec. 20' 



£ = 

P 

2S = 
8 = 



32° 20' 
43° 25' 



= m°zrw 

= 145° 22' 54" 
72° 41' 27" 

40° 21' 22" 



log sec = 
log esc = 

log cos = 
log sin = 



0.13884 
0.02805 

9.47353 
9.81126 



d. 

May 21 



2 )19.45168 
logsin^= 9.72584 
it=32°8' 6". 

h. m. s. 



19 42 55 
3 35 



May 21 19 39 20 

May 21 20 59 14 

1 19 54 

19° 58' 30" 



Local apparent astronomical time. 

Equation of time. 

Local mean astronomical time. 

Greenwich mean time. 

Difference in time. 

W. Longitude. 




Equation of time. 



3 34.66 

0.55 

35.21 



£ = 64° 16' 12" = 4h 



8. 1895, Aug. 24, a.m., at ship, latitude at noon /37° 59' N. ; observed 
altitude Q_ 37° 13' 30"; index correction + 2' 44"; height of eye 18 ft. 
time by chronometer Aug. 23 d. 18h. 13m. 24 s., which was 1 m. 5s. fast 
for mean noon at Greenwich, August 1, and on August 10 was m. 42 s. 
slow for mean time at Greenwich; course (true) since observation N.N.W. ; 
distance 22.4 miles. Required the longitude at noon. 
Chronometer. 

d. h. 

Aug. 1 fast 1 5 Aug. 23 18 13 24 2/ = 37° 59' 0" ; 

Aug. 10 slow 42 + 42 L d = 20' 41' 

9 )1 47 + 2 44 



11.89 

13.76 

2 43.60 



Aug. 23 18 16 50 



L = 37° 38' 19" N. 



teachers' edition. 



379 



37° 13' 


30" 


' Index cor. , 


+ 2' 44" 


G'sdec 


. 11° 6'46.7"N. 


51.38 




Semi-diam. , -f 15' 52" 
Dip, - 4' 9" 




4fB8.fr 


5.72 


+ 13' 17"- 


ii°ii'40.6"N. 


293.89 




Refraction, — 1" 17" 




90° N. 




Parallax, + 0' 7" 


v- 


= 78° 48' 19" N. 


h= 37° 26' 47" 




L= 37° 38' 19" log sec = 0.10134 


Equation of time. 


p = 78° 48 / 19" 


log esc = 0.00834 




m. s. 


2S= 153°53 / 25 ,/ 


2 16.42 


0.661 


8 = 76°56 / 42 // log cos = 9.35389 




3.78 


5.72 


R = 39° 29' 55" log sin = 9.80350 


2 20.20 


3.78 


2)19.26707 




logsini*= 9.63353 




i^ = 25°28 / 18". 




« = 50° 56' 36" = 3 h. 22 


m. 46 s. 


Mer.Z d =26.1 


log= 1.41664 


(1. h. m. s. 

Aug. 23 20 36 14 


Local appar. ast. time. 


C=22 o 30' 


logl 


tan=9.61722 


2 20 


Equation of time. 


log 1 


\ a =1.03386 


Aug. 23 20 38 34 


Local mean ast. time. 


Xd= 10.811 


Aug. 23 18 16 50 


Greenwich mean time. 


= 10'49"W. 


2 21 44 


Difference in time. 


35° 26' 0" 


E. Long, at sights. 










L0' 49" 


W. Long, since 


observ. 



35° 15' 11" E. Loner, at noon. 



9. 1895, Jan. 29, p.m., at ship, latitude 28° 45" X.; observed altitude 
17° 46" 30"; index correction — 3' 18"; height of eye 16 ft., time by 
chronometer January 28 d. 16 h. 31m. 30 s., which was 1 m. 16.5 s. fast 
for Greenwich mean noon, December 17, 1894, and on January 1, 1895, 
was 1 m. 3 s. slow for Greenwich mean time ; course (true) since noon 
N.W. by W. ; distance 20 miles. Required the longitude at the time of 
observation, and also at noon. 



Chronometer. 




m 


s. 


, Dec. 17 fast 1 


16.5 


, Jan. 1 slow 1 


3. 


15)2 


19.5 




9.30 




28.02 



Jan. 28 16 31 30 

+ 1 3 

+4 21 

Jan. 28 16 36 54 



4 20.59 



380 



NAVIGATION. 



17° 46' 30" 



+ 6 7 11'- 



h = 


17° 


52' 


41" 


L = 


28° 


45' 


0" 


P = 


108° 


cr 


5" 



'Index cor., 
Semi-diam. , 
Dip, 

Refraction, 
Parallax, 

log sec = C 
log esc = ( 



2.8 = 154° 37 / 46 

S= 77° 18' 53' 
B= 59° 26' 12 



log cos = 9. 

log sin — 9. 

2 )19, 

log sin \ t = 9. 

i* = 28°26'18". t = 



- 3' 18' 
+ 16' 16' 

- 3' 65' 

- 3' 0' 
+ / 8' 

.05714 
..02179 

.34163 
93504 



O's dec. 17° 55' 7.1* 
4' 58.4" 



18° 
90° 



0' 5.5' 



40.43 

7.38 



298.37 



p = 108° 0' 5.5"K 
Equation of time. 

m. s. 

13 20.81 

3.21 

13 17.60 



0.435. 

7.38 



3.21 



35560 



Z«*=11'7" 
Mer.i d =12.6 

0=56° 15' 



log= 1.10037 

log tan= 10.17511 

\og\ d = 1.27548 

\ d = 18.857 

= 18 / 51 // E. 



67780 
56°52 / 36 // =3h. 

d. h. m. s. 

Jan. 29 3 47 30 
13 18 



Jan. 29 4 48 
Jan. 28 16 36 54 



11 23 54 

170° 58' 30" 

18 / 51" 

171° 17 / 21 // 



47 m. 30 s. 

Local appar. ast. time. 
Equation of time. 
Local mean ast. time. 
Greenwich mean time. 
Difference in time. 
E. Long, at sights. 
W. Long, since noon. 
E. Long, at noon. 



10. 1895, Aug. 31, p.m., at ship, latitude 0°; observed altitude 45° 
5' 30"; index correction — 2' 4"; height of eye 15 ft. ; time by chronom- 
eter Aug. 31 d. 9h. 11m. 28 s., which was 5 m. 20 s. fast for mean noon 
at Greenwich April 15, and on June 16 was fast 2 m. 43 s. on mean time 
at Greenwich. Required the longitude. 
Chronometer. 

d. h. m. s. 



April 15 fast 5 20 






Aug. 31 9 11 28 




June 16 fast 2 43 






-2 43 




62)2 37 






+ 3 12 




2.53 


Aug. 31 9 11 57 




76 








3 12.28 








45° 5' 30" 


r Index cor. , 


- 2' 4" 


0'sdec. 8°38 / 56.1"N. 


54.11 




Semi-diam., 
Dip, 


+ 15' 53" 
- 3' 48" 


8 / 17.8" 


9.20 


+ 9' 9" \ 


8° 30' 38.3" N. 


497.81 




Refraction, 


- 0'58" 


90° N. 






Parallax, 


+ 0' 6" 


p - 81° 29' 22" N. 




= 45° 14' 39" 





TEACHERS' EDITION. 



381 



h= 45° 14' 39" 
L= 0° 0' 0" 
p= 81° 29' 22" 



2 S = 126° 44' 
S = 
R = 



1" 

63° 22' 0" 
18° 7' 21" 



log sec — 
log esc = 

log cos = 
log sin = 



0.00000 
0.00481 

9.65155 

9.49283 



Equation of time. 

m. b. 

15.51 
7.11 

8.40 



0.773 
9.20 



7.11 



2 )19.14919 
log sin -H= 9.57459 

i t = 22° 3' 15" 

d. h. m. s. 



t - 44° 6' 30" = 2 h. 56 m. 26 s. 



Aug. 31 2 56 26 

8 

Aug. 31 2 56 34 

Aug. 31 9 11 57 

6 15 23 

93° 50' 45" 



Local apparent astronomical time. 

Equation of time. 

Local mean astronomical time. 

Greenwich mean time. 

Difference in time. 

W. Longitude. 



11. 1895, April 15, a.m., at ship, latitude 48° 52' N. ; observed altitude 
22° 18' ; index correction — 3' 54" ; height of eye 17 ft. ; time by chro- 
nometer April 14 d. 22 h. 30 m. 42 s., which was m. 4 s. slow for mean 
noon at Greenwich January 1, and on January 12 was fast m. 2 s. 
Required the longitude. 
Chronometer. 

m. s. 

Jan. 1 slow 4 
Jan. 12 fast 2 



11 )0 6 



0.545 
93 



d. h. m. s. 

Apr. 14 22 30 42 
-0 2 

- 51 

Apr. 14 22 29 49 



22° 18' 0" 



+ 5' 49" 



50.69 
r Index cor. , — 3' 54' 
Semi-diam., -f- 15' 58' 
Dip, — 4' 2 / 

Refraction, — 2' 21' 
Parallax, -f 0' 8' 



0's dec. 



9°46 / 35.5" 
1'20.4" 



N. 



9° 45' 15. V 

90° 



53.58 
1.50 



80.37 



# = 80° 14' 45" N. 



L- 48°52 / 0" log sec ■= 0.18190 


Equation of time. 




p = 80° 14' 45" log esc = 0.00632 


m. s. 




2 6 = 151° 30' 34" 


1.59 


0.619 


S= 75°45 / 17" log cos = 9.39107 
R= 63°21 , 28 // log sin = 9.90438 


0.93 
2.52 


1.50 


0.93 


2)19.48367 






logsin£*= 9.74183 






£* = 33°29'41". 


t = 66° 59' 22" = 4 h. 27 1 


q. 57 s. 



382 



NAVIGATION. 



d. h. m. 8. 

April 14 19 32 3 

3 

April 14 19 32 6 

April 14 22 29 49 

2 57 43 

44° 25' 45" 



Local apparent astronomical time. 

Equation of time. 

Local mean astronomical time. 

Greenwich mean time. 

Difference in time. 

W. Longitude. 



12. 1895, Aug. 28, p.m., at ship, latitude 5° S. ; observed altitude 
38°; index correction + 5 / 27"; height of eye 21 ft. ; time by chronom- 
eter Aug. 27 d. 22 h. 20 m. 30 s., which was 10 m. s. slow for mean noon 
at Greenwich Feb. 19, and on May 30 was 2 m. 20 s. slow on mean noon 
at Greenwich. Kequired the longitude. 
Chronometer. 



d. h. m. s. 



Feb. 19 slow 10 
May 30 slow 2 20 



Aug. 27 



38° 0' 0" 



+ 15' 44' 



100 )7 40 

4.6 

90 

6 54 

f Index cor., + 5' 27" 
Semi-diam., + 15' 53" 
Dip, - 4' 29" 

Refraction, — V 14" 
Parallax, + 0' 7" 



22 20 30 
+ 2 20 
-6 54 



Aug. 27 22 15 56 



O's dec. 



9° 43' 13.4" 

rsi.7" 



N. 



53.02 
1.73 



h = 


38° 15' 44" 


L = 


5° 0' 0" 


P = 


99° 44' 45" 



2 S = 143° 0' 29" 
S= 71° 30' 14" 
B= 33° 14' 30" 



log sec = 
log esc = 

log cos = 
log sin = 



0.00166 
0.00632 

9.50139 
9.73891 



9°44 / 45.1"N.| 91.72 

90^ S. 

p=r99°44 / 45" S. 

Equation of time. 

m. s. 

1 9.63 0.729 

1.26 1.73 



1 10.89 



1.26 



2 )19.24828 
logsini£= 9.62414 
i* = 24° 53' 20" 
t = 49° 46' 40 A 



= 3h. 19m. 7s 



Aug. 



d. 

28 



19 7 Local apparent astronomical time. 

1 11 Equation of time. 

Aug. 28 3 20 18 Local mean astronomical time. 

Aug. 27 22 15 56 Greenwich mean time. 

5 4 22 Difference of time. 

76° 5' 30" E. Longitude. 



TEACHERS' EDITION. 



383 



13. 1895, Sept. 22, a.m., at ship, on the equator, observed altitude 
17° 20" 40"; index correction — V 9"; height of eye 20 ft.; time by 
chronometer Sept. 22 d. 4h. 59 m. 16 s., which was 15 s. slow for Green- 
wich mean noon, April 30, and on June 1 was 10.6 s. fast for mean time 
at Greenwich. Required the longitude. 



d. h. m. 8. 

Sept. 22 4 59 16 
-0 10.6 
- 1 30.6 



Chronometer 






m. 


8. 


April 


30 slow 


15 


June 


1 fast 


10.6 




32)0 


25.6 







0.8 
113.2 



Sept. 22 4 57 35 



17° 20' 40" 



- 24' 37' 



h = 


16° 56' 3" 


L = 


0° 0' 0" 


v- 


89° 46' 9" 



1 30.6 

Index cor. , — V 19" 
Semi-diam., — 15' 59" 
Dip, - 4' 23" 

Refraction, — 3' 4" 
Parallax, 4- 0' 8" 



O'sdec. 0° 



18' 41. 2" X. 
4' 50.1" 



58.48 
4.96 



0° 13' 51. r 

90° 



1290.06 



2 8 = 106° 42' 12" 






53° 21' 
36° 25' 



log sec = 
log esc = 

log cos = 
log sin 



0.00000 
0.00000 

9.77590 
9.77354 



p = 89°46' 9" X. 
Equation of time. 

m. s. 

7 15.58 
4.32 



7 19.90 



0.807 
4.96 



4.32 



2 )19.54944 
log sin ^ = 9.77472 

-H = 36° 31' 57". 
t - 72° 3' 54" = 4 h. 48 m. 16 s. 



Sept. 



d. 

21 



Sept. 21 19 4 24 

Sept. 22 4 57 35 

9 53 11 

148° 17' 45" 



Local apparent astronomical time. 

Equation of time. 

Local mean astronomical time. 

Greenwich mean time. 

Difference in time. 

W. Longitude. 



14. 1895, Aug. 5, a.m., at ship, latitude at noon 30° 30' N. ; observed 
altitude 35° 6'; height of eye 15 ft. ; time by chronometer 5 d. 8 h. 
39 m. 22 s., which was fast 29 m. 32.4 s. on Greenwich mean noon, July 8, 
and on July 20 was fast 30 m. s. on Greenwich mean noon ; course 
(true) till noon W.; distance 48 miles. Required the longitude at 
noon. 



384 



NAVIGATION. 



Chronometer. 

m. s. 

July 8 fast 29 32.4 

July 20 fast 30 0.0 

12 )0 27.6 

2.3 

16.4 

37.7 



(1. h. m. s. 

Aug. 5 8 39 22 
-30 
- 38 

Aug. 5 8 8 44 



35° 6' 0" 



+ 10' 45" 



h = 


35°16 , 45 // 


L = 


30° 30' 0" 


P = 


73° 6' 11" 



Index cor., 4- 0" 0" 
Semi-diam., + 15" 49' 
Dip, - 3' ±8' 

Refraction, - V 23' 
Parallax, + 0' T 



0's dec. 16 c 



59' 19.9' 
5'30.9 A 



N. 



16° 53' 49.0" 
90° 



40.60 
8.15 



330.89 



2S= 138° 52' 56" 
8 = 69° 26' 28" 
B= 34° 9' 43" 



log sec : 

log CSC : 

log COS = 
log sin =_ 



0.06468 
0.01916 

9.54552 
9.74938 



2 )19.37874 
logsin|£= 9.68937 
it = 29° 16' 46" 



p = 73° 6' 11" 
Equation of time. 

m. s. 

5 48.79 
2.04 



N. 



5 46.75 



0.250 
8.15 



2.04 



t = 58° 33' 32" = 3 h. 54 m. 14 s. 



Aug. 



h. 

20 



5 46 
5 47 



Aug. 4 20 11 33 

Aug. 5 8 8 44 

11 57 11 

179° 17' 45" 

48 miles = 55 7 43" 

180° 13 / 28" 

= 179° 46' 32" 



Local apparent astronomical time. 

Equation of time. 

Local mean astronomical time. 

Greenwich mean time. 

Difference in time. 

W. Longitude at sight. 

W. 

E. Longitude at noon. 



15. 1895, Nov. 12, a.m., at sea, in latitude 7° 10' N. ; four observed 
altitudes of the were taken at the times (by watch) standing opposite, 



Obs. alt. 21° 8' 40" 


2 55 48 


11' 50" 


56 


14' 50" 


56 13 


17' 30" 


56 26.5 



TEACHERS* EDITION. 



385 



Index correction + 31"; height of eye 18 ft.; correction of watch by 
chronometer — 5h. 12 m. 2.1 s. Required the longitude. 







h. m. s. 










21° 8' 40" • 




2 55 48 


O's declination. 






ir 50" 




06 


17° 43' 11.1" 


S. 




40.67 


14' 50" 




56 13 
56 26.5 


V 33.3" 






2.27 


IV 30" 


17° 41' 87.8" 


S. 


93.32 


52' 50" 




224 27.5 


90° 


X. 






21° 13' 12.5" 




2 56 6.9 
-5 12 2.1 


p = 107° 41' 38" 


N. 










21 44 4.8 


Index correction, 




4- 


0' 31" 








Semi-diameter, 




+ 16' 12" 








Dip, 




— 


4 9" 








Refraction, 




— 


2' 29" 








Parallax, 




— 


0' 8" 


+ 10 / 13" 


+ 10' 13" 


h= 21°23 / 25" 














L= 7° 10' 0" 


log sec = 0.00341 


Equation of time. 






p = 107° 41 / 38" 


log esc == 0.02104 


m. s. 








2 S = 136° 15' 3" 






15 44.88 






0.324 


S= 68° 7' 31" 


log cos = 9.57122 


0.74 






2.27 


R= 46° 44' 6" 


log sin = 9.86224 


15 45.62 






0.74 






2)19.45791 










log sin 


%t= 9.72895 














|* = 32° 23' 38". 














t = 64° 47' 16" = 


= 4h. 19 m. 9 s. 






' 


d. 


h. m. 


s. 










Nov. 11 


19 40 


51 Local appare 


nt astronomical time 








15 46 Equation of 


time. 








Nov. 11 


19 25 


5 Local mean 


astronomical time. 








Nov. 11 


21 44 

2 19 

34° 45 y 


5 Greenwich n 

Difference ir 

0" W. Longituc 


lean time. 
l time, 
le. 









16. 1895, Nov. 13, a.m., at sea, in latitude 9° 30' N.; five observed 
altitudes of the were taken at the times (by watch) standing opposite, 





h. m. s. 


. alt. 18° 58' 40" 


2 59 2 


19° T20" 


13 


3' 30" 


28 


r 30" 


45 


ir 0" 


57.5 



386 



NAVIGATION. 



Index correction -f 32' 
chronometer — 5 h. 11m. 



; height of eye 18 ft. ; correction of watch by 
58.4 s. Required the longitude. 



© 


18° 58' 


40" 


h. m. s. 

2 59 2 


0's declination. 








19° V 


20" 


13 


17° 59' 18.1" 


S. 




39.90 




3' 

r 


30" 
30" 


28 
45 


V 28.2" 






2.21 




17° 57' 49.9" 


8. 


88.18 




ii' 


0" 


57.5 


90° 


N. 








22' 


0" 


145.5 


p = 107° 57' 50" 


N. 








19° 4' 


24" 


2 59 29.1 
















-5 11 58.4 


Index correction, 




+ 


0' 32" 








21 47 30.7 


Semi-diameter, 




+ 16' 13" 










Dip, 




— 


4' 9" 










Refraction, 




— 


2' 48" 










Parallax, 




+ 


0' 8" 




+ 9' 56" 








+ 


9 7 56" 


h = 


19° 14' 


20" 












Z = 


9°30 / 


0" 


log sec = 0.00600 


Equation of time. 






v- 


107° 57' 


50" 


log esc = 0.02171 


m. s. 








2S = 


136° 42' 


10" 




15 36.66 






0.361 


S = 


68° 21' 5" 
49° 6' 45" 


log cos = 9.56692 

log sin = 9.87852 


0.80 
15 37.46 






2.21 


R = 


0.80 








2)19.47315 














logsini*= 9.73657 
















££=33° 2' 22". 
















t = 66° 4' 44" = 


4h. 24 m. 19 s. 












d. , 


h. m. s. 












Nov 


. 12 


19 35 41 Local appar* 


mt astronomical time 


. 










15 37 Equation of 


time. 










Nov 


. 12 


19 20 4 Local mean 


astronomical time 










Nov 


. 12 


21 47 31 Greenwich n 
2 27 27 Difference ii 


lean time. 
i time. 












36° 5r 45" W. Longituc 


le. 









17. 1895, Nov. 17, a.m., at sea, in latitude 15°35'N.; five observed 
altitudes of the were taken at the times (by watch) standing opposite, 
viz. : 

Obs. alt. 23° 56' 0" 

24° 0' 0" 

4' 0" 

6' 10" 

10' 0" 



h. 


m. s. 


4 


12 31 




56 




13 2.5 




14 




28.5 



teachers' edition. 



387 



Index correction 4- 31"; height of eye 18 ft.; correction of watch by 
chronometer — 5h. 11 m. 43.6 s. Required the longitude. 









h. m. s. 













23° 56' 0" 




4 12 31 


O's declination. 








24° 0' 0" 




46 


19° / 34.2 // 


8. 




36.63 




4' 0" 
6' 10" 




13 2.5 
14 


35.9" 






0.98 




18° 59' 58.3" 


S. 


35.90 




10' 0" 




28.5 


90° 


N 








16' 10" 




65 2 


p = 108° 59' 58" 


N. 








24° 3' 14" 




4 13 0.4 
















-5 11 43.6 


Index correction, 




+ 


0' 31" 








23 1 16.8 


Semi-diameter, 




4- 16' 13" 










Dip, 




- 


4' 9" 










Refraction, 




-r 


2' 10" 










Parallax, 




4- 


0' 8" 




+ 10' 33" 










4- 10' 33" 


h = 


24° 13' 47" 














L~ 


15°35 / 0" 


log 


sec = 0.01627 


Equation of time. 






P = 


108° 59' 58" 


log 


esc = 0.02433 


m. s. 








2S = 


148° 48' 45" 






14 55.18 






0.503 


S = 


74°24 / 22" 


log cos = 9.42946 


0.49 






0.98 


R = 


50° 10' 35" 


logs 


mi = 9.88537 
2)19.35543 


14 55.67 






0.49 




log sin 


it= 9.67771 
















it = 28° 25' 55". 
















t = 56° 51' 50" = 


= 3h. 47 m. 27 s. 










d. 


h. m 


s. 












Nov. 16 


20 12 


33 Local appare 


nt astronomical time 










14 


56 Equation of 


time. 










Nov. 16 


19 57 


37 Local mean i 


astronomical time 










Nov. 16 23 1 


17 Greenwich n 


Lean time. 












3 3 40 Difference in 


time. 










45° 55' 


0" W. Longitud 


e. 









18. 1895, Nov. 18, a.m., at sea, in latitude 16° 25' N.; five observed 
altitudes of the were taken at the times (by watch) standing opposite, 



Obs. alt. 





h. 


m. s. 


° 13' 30" 


3 


52 42 


16' 10" 




53.5 


19' 20" 




53 6.5 


22' 30" 




23 


25' 30" 




38 



388 



NAVIGATION. 



Index correction + 32"; height of eye 18 ft.; correction of watch by 
chronometer — 5h. 11 m. 39.9 s. Required the longitude. 



18° 13' 30" 
16' 10" 
19' 20" 
22' 30" 
25 / 30" 
97' 0" 
18° 19' 24" 



+ 7 51" 

7i = 18° 29' 15" 

L= 16° 25' 0" 

p = 1090 14' 16" 

2S = 1440 8 / 3r 



& = 
22 = 



72° 4'15 A 
53° 35' A 



3 52 42 
53.5 
53 6.5 
23 

38 



265 43 
3 53 8.6 
-5 11 39.9 

22 41 28.7 



log sec = 
log esc = 

log cos = 
log sin = 



0.01808 
0.02493 

9.48832 
9.90565 



2 )19.43698 
logsin££= 9.71849 
i* = 81° 31'- 57' 
t = 63° 3' 54" 



0's declination. 
19° 15' 3.2" S. 

46.9" 
19° 14' 16.3" S. 
90° N. 

p - 109° 14 / 16" N. 

Index correction, + 
Semi-diameter, -f 

Dip, 

Refraction, — 

Parallax, -f 

■f 

Equation of time. 

m. s. 

14 42.70 
0.70 



35.77 
1.31 



46.86 



0'32 
16 / 14 
4 
2' 54 

0' 8" 



14 43.40 



= 4h. 12 m. 16s. 



d. 
Nov. 17 



19 47 44 Local apparent astronomical time. 

14 43 Equation of time. 

Nov. 17 19 33 1 Local mean astronomical time. 

Nov. 17 22 41 29 Greenwich mean time. 

3 8 28 Difference in time. 

470 r // w Longitude. 



\ 



0.537 

1.31 

0.703 



19. 1895, Dec. 4, a.m., at sea, in latitude 36°10 / N.; five observed 
altitudes of the were taken at the times (by watch) standing opposite, 
viz.: 

h. m. s. 

Obs. alt. 13° / 30" 6 

3' 10" 

5' 40" 

8' 50" 

12' 0" 



27 14 
29.5 
49 
5 
23 



28 



teachers' edition. 



389 



Index correction + 32"; height of eye 18 ft.; correction of watch by 
chronometer — 5h. 10 m. 47.1s. Required the longitude. 



II 



13 c 



0' 


30" 


3' 


10" 


5' 


40" 


8 / 


50" 


12' 


0" 


30' 


10" 


6' 


2" 



h. 


m. 


s. 


6 


27 


14 

29.5 

49 




28 


5 
23 





139 


0.5 


6 


27 


48.1 


—5 


10 


47.1 



1 17 1 





rf 


8' 43" 


h = 


13° 


14' 


45" 


L = 


36° 


10' 


0" 


P = 


112° 


16' 


3" 



2 S = 161° 40' 48" 
5= 80° 50' 24" 
E= 67° 35' 39" 



log sec = 
log esc = 

log cos = 
log sin = 



0.09296 
0.03366 

9.20192 
9.96591 



O's declination. 
22° 15' 37.3" S. 
25.7" 



22° 16' 
90° 



3.0" 



S. 
N. 



20.10 
1.28 



25.73 



p=112°16' 3" N. 



Index correction, + 0' 32" 



Semi-diameter, 
Dip, 

Eefraction, 
Parallax, 



4- 16' 16" 

- 4' 9" 

- 4' 5" 
+ 0' 9" 
+ 8'43 A 



2 )19.29445 
logsini« = 9.64722 

i * = 26° 20' 55" 
* = 52° 41' 50" 



Equation of time. 



9 40.05 
1.30 

9 38.75 



= 3 h. 30 m. 47 s. 



d. h. m. b. 

Dec. 3 20 29 13 
9 39 



Local apparent astronomical time. 
Equation of time. 
Dec. 3 20 19 34 Local mean astronomical time. 
Dec. 4 1 17 1 Greenwich mean time. 
4 57 27 Difference in time. 
74° 21' 45" W. Longitude. 



1.013 

1.28 



1.30 



20. 1895, Dec. 4, p.m., at sea, in latitude 36° 36' N.; four observed 
altitudes of the were taken at the times (by watch) standing opposite, 



h. m. s. 



alt. © 16° 31' 10" 


1 7 26.5 


30' 0" 


35.5 


28' 30" 


46 


27' 20" 


56.5 



^o 



390 



NAVIGATION. 






Index correction 
chronometer — 5 h. 



+ 30"; height of eye 18 ft.; correction of 
10 m. 46.1 s. Required the longitude. 



watch by 







h. m. s. 










16°31 / 10 // 




1 7 26.5 


O's declination. 






30' 0" 




35.5 


22°15 / 37.3 / 


'S. 




20.10 


28' 30" 




46 
56.5 


2' 39.8' 






7.95 


27' 20" 


22° 18' 17. V 


'S. 


159.79 


117' 0" 




164.5 


90° 


N. 






16° 29' 15" 




1 7 41.1 
-5 10 46.1 


p = 112° 18' 17" 


N. 










7 56 55 


Index correction, 




+ 


0' 30" 








Semi-diameter, 




+ 16' 16" 








Dip, 




— 


4' 9" 








Refraction, 




— 


3' 14" 








Parallax, 




+ 


0' 8" 


+ 9 / 31 // 










+ 


9' 31" 


h= 16° 38' 46" 














L= 36° 36' 0" 


log 


sec= 0.09538 


Equation of time 






p = 112° 18' 17" 


log 


csc= 0.03377 


m. s. 








2 8 = 165° 33' 3" 






9 40.05 






1.013 


8= 82° 46' 31" 


log 
log 


cos = 9.09955 
sin = 9.96116 


8.05 
9 32.00 






7.95 


#= 66° 7' 45" 


8.05 






2)19.18986 










log sin 


it= 9.59493 














i t = 23° 10' 18". 














t - 46° 20' 36" = 


= 3h. 5m. 22s. 








d. 


h. m 


. s. 










Dec. 4 


3 5 22 Local appare 


tit astronomical time. 








9 32 Equation of t 


ime. 








Dec. 4 


2 55 


50 Local mean a 


LStronomical time. 








Dec. 4 


7 56 
5 1 


55 Greenwich m 
5 Difference in 


ean time, 
time. 








75 


°16 / 


15" W. Longitud 


e. 










: "^o* 



2, ■ * o , a. "* <$F Qj 





-o- tv 




"^ 0* 























%o^ * 







^o* 



% 






&><< 

<£ ^ 






"^0* 


















p y ° * >> 

% 



J % 













*~\.A* - 







^ 















4 .^ 































'^o x 



\o^ 







■ 



Hon 



>**«:»! 



: V;H 



T.i«l 



